Eastern Cape2025MarchTestMemo foundArithmeticQuadraticGeometricSigma
2025 G12 CT March EC Amathole District
Source folder: Tests
Source file: 2025 G12 CT March EC Amathole District.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 1 1.1 Given the quadratic pattern -80 ; -63 ; -48; ... ... 1.1.1 Write down the next two terms of the pattern. (1) 1.1.2 Determine the general term of the pattern in the form of Tn= an2 + bn+ c (3) 1.1.3 Between which two consecutive terms of the quadratic pattern will the first difference be -103 ? (3) 1.2 Consider the following sequence : 16(p-3)3 ; 8(p-3)4 ; 4(p-3)5; ... ... . p≠3 1.2.1 For which values of p will the series converge? (3) 1.2.2 Calculate the sum to infinity if p= 3,5 (2) [12] [Page 4] QUESTION 2 2.1 The sum of the first n terms of a arithmetic series is given by: Sn= 4n2 + 6n 2.1.1 Determine the first three terms of the series. (2) 2.1.2 The last term of the series is 58 , how many terms are in the series. (3) 2.1.3 Write down the series in sigma notation. (2) 2.2 The sum of the first 5 terms of a convergent geometric series is 62 and the sum to infinity of the series is 64. Determine the common ratio. (4) 2.3 KHANYA FM a community radio in Butterworth had a competition where the prize money is awarded over a period of 5 days. On the first day R5 000 is given to the winner. On the second day, 80% of that prize money was awarded to the winner and so on such the prize money continued to be 80% of the amount awarded on the previous day is given to the winner of the day. 2.3.1 How much money i
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Matched memo: 2024 G12 CT March EC Amathole East Memo.pdf
Relevant memo page(s): 2
[Page 2] a b b a a a n 2 a 0 b 1 c n T (4) 1.1.3 Minimum value. 0 a min Reason (2) 1.2 45 3 1 1 30 1 r a S 3 1 r correct substitution answer (3) 1.3 6 4 1 5 1 1 1 p p r convergent condition r answer (3) 1.4 99977112 , 59 2 1 1 2 1 1 30 1 1 18 18 r r a S 51 17 3 17 S 111 9977112 . 110 17 18 35 S S S correct substitution 18 S 17 S answer (4) [18]
Free State2025MarchTestAfrikaansMemo foundQuadraticGeometricConvergence
2025 G12 CT March Free State Afrikaans
Source folder: Tests
Source file: 2025 G12 CT March Free State Afrikaans.pdf
Relevant question page(s): 4
[Page 4] VRAAG 2 2.1 Gegee die meetkundige reeks: 2 3 4 8 4 2 ....... x x x - + + 2.1.1 Vir watter waarde(s) van x sal die reeks konvergeer? (3) 2.1.2 Bepaal die Sas x = 3 2 (2) 2.2 Die tweede term van 'n kwadratiese getalpatroon is 13. Die algemene term van die eerste verskil van die kwadratiese getalpatroon word gegee deur 8 2 nT n = + . 2.2.1 Bepaal die eerste 5 terme van die kwadratiese getalpatroon. (3) 2.2.2 Bepaal dus die algemene term 2 nT an bn c = + + (4) 2.3 Gegee: 2 ; 3 ; 5 ; 6 ; 8 ; 12... 2.3.1 Bepaal die volgende twee terme in die ry. (2) 2.3.2 Bepaal die som van die eerste 39 terme. (5) 2.4 Skryf die volgende reeks in notasie 64 32 16 8. ... 5 5 5 5 - + - + tot n terme. (3) 2.5 Die getalle 7 ; x ; y vorm 'n meetkundige ry. As die som van hierdie drie getalle (-2x - 7) is, bepaal die moontlike waarde(s)
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Matched memo: 2024 G12 CT March Free State Memo.pdf
Relevant memo page(s): 3
[Page 3] 3 1.5 (√ab 3 ) 1 2. (b√a 3 ) 1 b [(ab) 1 3] 1 2 . [b(a 1 3)] 1 b a 1 6. b 1 6. b 1 b. a 1 3b a 3b+6 6.3b. b 1 6+1 b a b+2 6b. b b+6 6b (√ab 3 ) 1 2. (b√a 3 ) 1 b a 1 6. b 1 6 b 1 b. a 1 3b a 3b+6 6.3b. b 1 6+1 b (4) [20] QUESTION 2 2.1.1 25 16 ; 81 16 ; 169 16 ; 289 16 7 2 11 2 15 2 2 2 Second constant difference = 2 (1) 2.1.2 2a= 2 a= 1 3a+ b= 7 2 3(1) + b= 7 2 b= 1 2 a+ b+ c= 25 16 1 + 1 2 + c= 25 16 c= 1 16 ∴Tn= n2 + 1 2 n+ 1 16 a= 1 b= 1 2 c= 1 16 (3) 2.1.3 ∴Tn= n2 + 1 2 n+ 1 16 Tn= (n+ 1 4) (n+ 1 4) Tn= (n+ 1 4) 2 (n+ 1 4) 2 > 0 and is a perferct square for n∈ N Factorisation Tn= (n+ 1 4) 2 Conclusion (3)
Free State2025MarchTestEnglishMemo foundQuadraticGeometric
2025 G12 CT March Free State English
Source folder: Tests
Source file: 2025 G12 CT March Free State English.pdf
Relevant question page(s): 4
[Page 4] QUESTION 2 2.1 Given the geometric series : 2 3 4 8 4 2 ....... x x x - + + 2.1.1 For which value(s) of x will the series converge? (3) 2.1.2 Determine the Sif x = 3 2 (2) 2.2 The second term of a quadratic number pattern is 13. The general term of the first difference of the quadratic number pattern is given by 8 2 nT n = + 2.2.1 Determine the first 5 terms of the quadratic number pattern. (3) 2.2.2 Hence determine the general term 2 nT an bn c = + + (4) 2.3 Given: 2 ; 3 ; 5 ; 6 ; 8 ; 12... 2.3.1 Determine the next two terms in the sequence. (2) 2.3.2 Determine the sum of the first 39 terms. (5) 2.4 Write the following series in notation: 64 32 16 8 ... 5 5 5 5 - + - + to n terms. (3) 2.5 The numbers 7 ; x ; y forms a geometric sequence. If the sum of these three numbers is (-2x - 7), determine the possible
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Matched memo: 2025 G12 CT March Free State Memo English.pdf
Relevant memo page(s): 4
[Page 4] 4 1.2 2 2 2 2 2 2 2 2 2 5 (2 5) 7 (4 20 25) 7 0 4 20 25 7 0 3 20 32 0 3 20 32 0 (3 8)( 4) 0 8 4 3 1 3 3 y x x x x x x x x x x x x x x x x or x y or y = - - - = - - + - = - + - - = - + - = - + = - - = = = = = 2 5 y x = - substitution factors values of x values of y (5) [20] QUESTION 2 2.1.1 1 1 1 1 2 2 2 r x x -< < -< - < -< < 1 1 2 x -< - < answer (3) 2.1.2 1 18 3 1 ( ) 4 72 10,29 7 a s r or ∞= - = -- = substitution into correct formula answer (2) 2.2.1 1st differences 1st term last 3 terms (3)
Free State2025MarchTestMemo foundQuadraticGeometric
2025 G12 CT March Free State
Source folder: Tests
Source file: 2025 G12 CT March Free State.pdf
Relevant question page(s): 4
[Page 4] QUESTION 2 2.1 Given the geometric series : 2 3 4 8 4 2 ....... x x x - + + 2.1.1 For which value(s) of x will the series converge? (3) 2.1.2 Determine the Sif x = 3 2 (2) 2.2 The second term of a quadratic number pattern is 13. The general term of the first difference of the quadratic number pattern is given by 8 2 nT n = + 2.2.1 Determine the first 5 terms of the quadratic number pattern. (3) 2.2.2 Hence determine the general term 2 nT an bn c = + + (4) 2.3 Given: 2 ; 3 ; 5 ; 6 ; 8 ; 12... 2.3.1 Determine the next two terms in the sequence. (2) 2.3.2 Determine the sum of the first 39 terms. (5) 2.4 Write the following series in notation: 64 32 16 8 ... 5 5 5 5 - + - + to n terms. (3) 2.5 The numbers 7 ; x ; y forms a geometric sequence. If the sum of these three numbers is (-2x - 7), determine the possible
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Matched memo: 2024 G12 CT March Free State Memo.pdf
Relevant memo page(s): 3
[Page 3] 3 1.5 (√ab 3 ) 1 2. (b√a 3 ) 1 b [(ab) 1 3] 1 2 . [b(a 1 3)] 1 b a 1 6. b 1 6. b 1 b. a 1 3b a 3b+6 6.3b. b 1 6+1 b a b+2 6b. b b+6 6b (√ab 3 ) 1 2. (b√a 3 ) 1 b a 1 6. b 1 6 b 1 b. a 1 3b a 3b+6 6.3b. b 1 6+1 b (4) [20] QUESTION 2 2.1.1 25 16 ; 81 16 ; 169 16 ; 289 16 7 2 11 2 15 2 2 2 Second constant difference = 2 (1) 2.1.2 2a= 2 a= 1 3a+ b= 7 2 3(1) + b= 7 2 b= 1 2 a+ b+ c= 25 16 1 + 1 2 + c= 25 16 c= 1 16 ∴Tn= n2 + 1 2 n+ 1 16 a= 1 b= 1 2 c= 1 16 (3) 2.1.3 ∴Tn= n2 + 1 2 n+ 1 16 Tn= (n+ 1 4) (n+ 1 4) Tn= (n+ 1 4) 2 (n+ 1 4) 2 > 0 and is a perferct square for n∈ N Factorisation Tn= (n+ 1 4) 2 Conclusion (3)
Gauteng2025MarchTestMemo foundArithmeticGeometricSigmaConvergence
2025 G12 CT March JHB North
Source folder: Tests
Source file: 2025 G12 CT March JHB North.pdf
Relevant question page(s): 3, 8
[Page 3] QUESTION 2 2.1 Prove that for any Arithmetic Sequence the sum of the first n terms is given by Sn= n 2[2a+ (n-1)d], if the first term is a and the common difference is d. (4) 2.2 Given the arithmetic series: 14 + 21 + 28 + 35 + ⋯+ 287. 2.2.1 How many terms are there in the series? (2) 2.2.2 Calculate the sum of all natural numbers from 12 to 115 that are NOT divisible by 7. (4) 2.3 Show that: ∑(6k-2) = 3n2 + n-4 n k=2 (3) [13] QUESTION 3 Given the geometric series: 8(p-3) + 4(p2 -9) + 2(p3 -3p2 -9p-27) + ⋯ 3.1 Determine the values of p for which the series will converge. (3) 3.2 If p= -2 , determine the sum to infinity of the series. (4) [7] [Page 8] right reserved Please turn over Mathematics NSC 2025 INFORMATION SHEET: MATHEMATICS x= -b± √b2 -4ac 2a A = P(1 + in) A = P(1 - in) A = P(1 + i)n A = P(1 - i)n ∑1 n i=1 = n ∑i n i=1 = n(n+ 1) 2 Tn= a + (n -1)d Sn= n 2 [2a + (n -1)d] Tn= arn-1 Sn= a(1-rn) 1-r ; r ≠1 S∞= a 1-r ; r ≠1 Fv = x[(1 + i)n -1] i PV = x[1 -(1 + i)-n] i f ′(x) = lim h→0 f(x + h) -f(x) h d = √(x2 -x1)2 + (y2 -y1)2 M( x1+x2 2 ; y1+y2 2 ) y = mx + c y -y1 = m(x -x1) m = y2 -y1 x2 -x1 m = tanθ (x - a) 2 + (y - b) 2 = r 2 In ∆ABC: a sinA= b sinB= c sinC a2 = b2 + c2 -2bc. cosA Area of ∆ABC = 1 2 ab. sinC sin( + ) = sin.cos + cos.sin sin( - ) = sin.cos - cos.sin cos( + )= cos.cos - sin.sin cos( - )= cos.cos + sin.sin cos2 = { cos2α-sin2α 2cos2α-1 1 -2sin2α sin2 = sin.cosα x̅ = ∑x n σ2 = ∑(xi-x̅)2 n i n P(A) = n(A) n((S) P(A or B) = P(A) + P(B
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Matched memo: 2025 G12 CT March JHB North Memo.pdf
Relevant memo page(s): 4
[Page 4] (29 11 ; 13 11) (1; 2) (6) [17] Q QUESTION 2 2.1 Sn = a+ (a+ d) + (a+ 2d) + ⋯+ l +(Sn= l+ (l-d) + (l-2d) + ⋯+ a) 2Sn = (a+ l) + (a+ l) + ⋯(a+ l) n times 2Sn = n(a+ l) Sn = n 2 (a+ l) But l= a+ (n-1)d Sn = n 2 [a+ a+ (n-1)d] Sn = n 2 [2a+ (n-1)d] ✓ Sn ✓ 2Sn= (a+ l) + ⋯(a+ l) ✓ 2Sn= n(a+ l) ✓ Substitution of l (4) 2.2.1 14 + 21 + 28 + 35 + ⋯+ 287 d= 21 -14 = 7 Tn = a + (n -1)d 287 = 14 + (n -1)(7) 7n-7 + 14 = 287 7n= 280 ∴n= 40 ✓ Substitution into the correct formula ✓ 40 (2)
General Source2025MarchTestMemo not linkedArithmeticQuadraticGeometricConvergence
2025 G12 March-Practice test with probable typo in 7.3
Source folder: Tests
Source file: 2025 G12 March-Practice test with probable typo in 7.3.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 1 5 ; 12 ; 21; 32;...is a Quadratic Sequence. 1.1 Write down the next term of the sequence. (1) 1.2 Determine the th n term of the sequence. (4) 1.3 Which term of the above sequence is 1152? (4) 1.4 Prove that none of the terms in sequence are perfect square (3) [12] QUESTION 2 2 ; 5 ; 8; 11;... is an Arithmetic Sequence. 2.1 Determine the first term that will be greater than 2012 (2) 2.2 Calculate the sum of the first 671 terms of the series (3) 2.3 If the Sum formula of the sequence: 2 ; 5 ; 8; 11;... is 3 1 2 n n S n : Determine the 12th term, by using the Sum formula. (4) [9] [Page 4] QUESTION 3 3.1 Prove that the sum to n terms of a Geometric series is given by 1 1 n n a r S r (4) 3.2 The first term of a geometric series is 12, the last term is 3 256 and the sum of the series is 6141 256 .Determine the common ratio and the number of term of the series. (6) [10] QUESTION 4 4.1 Find the value of 1 3 5 16 . p p (4) 4.2 For what values of x will the series 2 3 2(1 ) 4(1 ) 8(1 ) ... x x x be convergent? (3) [7]
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KwaZulu-Natal2025MarchTestMemo foundArithmeticQuadratic
2025 G12 CT March KZN
Source folder: Tests
Source file: 2025 G12 CT March KZN.pdf
Relevant question page(s): 3
[Page 3] QUESTION 1 1.1 Given: The sum to n terms of an arithmetic sequence is 2 3 5 n S n n = - . 1.1.1 Calculate the sum of the first 21 terms of this sequence. (2) 1.1.2 Determine the 22nd term of this sequence. (2) 1.1.3 How many terms of this sequence must be added to obtain a sum of 8162? (4) 1.2 Consider the sequence: 7 ; 7 ; 7 ; 12 ; 7 ; 17 ; 7 ; 22; ... 1.2.1 Determine the value of the 78th term of this sequence. (2) 1.2.2 Calculate the sum of the first 103 terms of this sequence. (4) [14] QUESTION 2 2.1 Given: ( ) 13 2 3 k k= - 2.1.1 Write down the values of the first three terms of the series. (2) 2.1.2 Write down the value of the constant ratio. (1) 2.1.3 Will ( ) 2 3 k k = - converge? Explain your answer. (2) 2.1.4 Calculate ( ) 13 2 3 k k x = - . Give your answer in terms of x. (3) 2.2 A quadratic sequen
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Matched memo: 2025 G12 CT March KZN Memo.pdf
Relevant memo page(s): 2
[Page 2] Mathematics Common Test March 2025 GRADE 12 Marking Guideline Copyright Reserved Please turn over 2 QUESTION 1 1.1.1 2 21 S 3(21) 5(21) 1218 = - = ✓A substitution ✓A answer (2) 1.1.2 ( ) ( ) 2 22 S 3 22 5 22 = - 1342 = 22 22 21 T S S = - 1342 1218 = - 124 = ✓A value of 22 S ✓CA answer (2) 1.1.3 2 8162 3 5 n n = - 2 3 5 8162 0 n n - - = ( ) ( ) ( )( ) ( ) 2 5 5 4 3 8162 2 3 n -- - - - = 154 53 or 3 n n = = - N/A 53 terms have to be added ✓A equating ✓A standard form ✓CA substitution ✓CA answer (53 only) (4) 1.2.1 The even-numbered terms form an AS with 7 a = and 5 d = . 39 T of the AS ( )1 a n d = + - ( ) 7 39 1 5 = + - 197 = ✓A substitution ✓CA answer (2) 1.2.2 Sum of the 52 odd-numbered terms 52 7 364 = = Sum of the 51 even-numbered terms ( ) 2 1 2 n a n d = + - ( ) ( ) 51 2 7 51 1 .5 2 = + - 6732 = Sum of f
KwaZulu-Natal2025MarchTestMemo not linkedArithmeticQuadraticGeometricConvergence
2025 G12 Feb_March Practice Test KZN
Source folder: Tests
Source file: 2025 G12 Feb_March Practice Test KZN.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 1 5 ; 12 ; 21; 32;...is a Quadratic Sequence. 1.1 Write down the next term of the sequence. (1) 1.2 Determine the th n term of the sequence. (4) 1.3 Which term of the above sequence is 1152? (4) 1.4 Prove that none of the terms in sequence are perfect square (3) [12] QUESTION 2 2 ; 5 ; 8; 11;... is an Arithmetic Sequence. 2.1 Determine the first term that will be greater than 2012 (2) 2.2 Calculate the sum of the first 671 terms of the series (3) 2.3 If the Sum formula of the sequence: 2 ; 5 ; 8; 11;... is 3 1 2 n n S n : Determine the 12th term, by using the Sum formula. (4) [9] [Page 4] QUESTION 3 3.1 Prove that the sum to n terms of a Geometric series is given by 1 1 n n a r S r (4) 3.2 The first term of a geometric series is 12, the last term is 3 256 and the sum of the series is 6141 256 .Determine the common ratio and the number of term of the series. (6) [10] QUESTION 4 4.1 Find the value of 1 3 5 16 . p p (4) 4.2 For what values of x will the series 2 3 2(1 ) 4(1 ) 8(1 ) ... x x x be convergent? (3) [7]
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Limpopo2025MarchTestAfrikaansMemo foundArithmeticGeometricSigmaConvergence
2025 G12 CT March Limpopo Afrikaans
Source folder: Tests
Source file: 2025 G12 CT March Limpopo Afrikaans.pdf
Relevant question page(s): 3, 4
[Page 3] VRAAG 2 2.1 Die volgende meetkundige reeks gegee: 2 24 12 6 3 ... x x x 2.1.1 Bereken die waarde van r , die gemene verhouding, in terme van x . (1) 2.1.2 As 4 x , bereken die som van die reeks vir die eerste 10 terme. (3) 2.2 Die som van die eerste n terme van 'n rekenkundige reeks word gegee deur 2 S 8 n n n 2.2.1 Bereken die som van die eerste 15 terme. (2) 2.2.2 Bereken die waarde van 15 T . (2) 2.2.3 As die eerste term van die ry 7 is, watter term van die ry sal 'n waarde hê van 169 ? (4) [12] [Page 4] VRAAG 3 3.1 Gegee dat: k= ∑(x-1)n ∞ n=1 3.1.1 Bereken die waardes van x vir wat k konvergeer. (2) 3.1.2 Bereken die waarde van k as 2. 3 x (4) 3.2 'n Lugverkeer kontrole toring met 11 horisontale balke word opgerig by 'n lughawe. Die vooraansig van die toring word in die diagram hieronder vertoon. Die lengte van die onderste horisontale balk is 70 cm lank en is vasgemaak op die grond. Elke addisionele balk is korter as die balk net onder dit. Die lengte van elk van die horisontale balke verminder met 'n konstante verskil van mekaar. As die boonste balk 50 cm is, watter balk sal 'n lengte hê van 58 cm? (5) [11]
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Matched memo: 2025 G12 CT March Limpopo Memo Afrikaans.pdf
Relevant memo page(s): 3
[Page 3] Wiskunde Graad 12 3 Limpopo DoE / Maart 2025 NSC- nasien riglyne Kopiereg voorbehou Blaai asseblief om VRAAG 2 2.1.1 3 2 T 6 T 12 2 x r x antwoord (1) 2.1.2 10 10 4 6 & 2 1 S 1 6 2 1 2 1 6138 n x a r a r r waardes van a en r vervanging antwoord (3) 2.2.1 2 15 S 15 8 15 105 vervanging antwoord (2) 2.2.2 15 15 14 T S S 105 84 21 metode antwoord (2) 2.2.3 1 1 2 2 2 n T S 7 14 21 S 2 8(2) 12 OR 7 14 21 T 5 14 28 2 2 T ( 1) 7 ( 1)( 2) a d d d d d a n d n 169 7 2 2 169 2 178 89 n n n OF n n-1 n 2 2 2 2 2 2 S S T 8 ( 1) 8( 1) 169 8 2 1 8 8 169 8 10 9 169 2 178 89 n n n n n n n n n n n n n n n 5 T 5 of 14 28 d 2 d vervanging antwoord formule vervanging vereenvoudiging antwoord (4) [12]
Limpopo2025MarchTestEnglishMemo foundArithmeticGeometricConvergence
2025 G12 CT March Limpopo English
Source folder: Tests
Source file: 2025 G12 CT March Limpopo English.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 2 2.1 Given the following geometric series: 2 24 12 6 3 ... x x x + + + + 2.1.1 Determine the value of r , the common ratio, in terms of x . (1) 2.1.2 If 4 x = , determine the sum of the series to 10 terms. (3) 2.2 The sum of the first n terms of an arithmetic series is given by 2 S 8 n n n = - + 2.2.1 Calculate the sum of the first 15 terms. (2) 2.2.2 Calculate the value of 15 T . (2) 2.2.3 If the first term of the series is 7 , which term of the series will have a value of 169 - ? (4) [12] [Page 4] QUESTION 3 3.1 n n 1 Given that: k ( 1) x = = - 3.1.1 Determine the values x for which k converges. (2) 3.1.2 Calculate the value of k when 2. 3 x = (4) 3.2 An air-traffic control tower with 11 horizontal supports is constructed at an airport. The front view of the tower is shown in the diagram below. The length of the bottom horizontal support is 70 cm long and is secured to the ground. Each additional support is shorter than the one below it. The length of each horizontal support decreases with constant difference from each other. If the top horizontal support is 50 cm, which horizontal support has a length of 58 cm? (5) [11]
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Matched memo: 2025 G12 CT March Limpopo Memo English.pdf
Relevant memo page(s): 3
[Page 3] Mathematics marking guideline 3 Limpopo DoE / March 2024 NSC Copyright reserved Please turn over QUESTION 2 2.1.1 3 2 T 6 T 12 2 x r x answer (1) 2.1.2 10 10 4 6 & 2 1 S 1 6 2 1 2 1 6138 n x a r a r r values of a and r substitution answer (3) 2.2.1 2 15 S 15 8 15 105 substitution answer (2) 2.2.2 15 15 14 T S S 105 84 21 method answer (2) 2.2.3 1 1 2 2 2 n T S 7 14 21 S 2 8(2) 12 OR 7 14 21 T 5 14 28 2 2 T ( 1) 7 ( 1)( 2) a d d d d d a n d n 169 7 2 2 169 2 178 89 n n n OR n n-1 n 2 2 2 2 2 2 S S T 8 ( 1) 8( 1) 169 8 2 1 8 8 169 8 10 9 169 2 178 89 n n n n n n n n n n n n n n n 5 T 5 or 14 28 d 2 d substitution answer formula substitution simplification answer (4) [12]
Mpumalanga2025MarchTestMemo foundArithmeticQuadraticGeometric
2025 G12 CT March Mpumalanga
Source folder: Tests
Source file: 2025 G12 CT March Mpumalanga.pdf
Relevant question page(s): 3
[Page 3] QUESTION 2 2.1 The given number pattern is a combination of a quadratic sequence and an arithmetic sequence : ;... 20 ; 20 ; 24 ; 12 ; 28 ;0 ; 32 ; 16 2.1.1 Determine the general term of the quadratic sequence (4) 2.1.2 Determine the general term of the arithmetic sequence (2) 2.1.3 The given number pattern has two consecutive terms that are equal in value, determine the positions of the two terms (4) 2.2 Calculate : 9 3 ) 3 ( 2 k k (4) [14] 1.1. Consider the arithmetic sequence: 8; 15; 22; ... 1.1.1 Determine the 36th term (2) 1.1.2 Calculate the sum of the first 36 terms (2) 1.1.3 If it is given that 786 72 72 m T T , determine the value ofm (3) 1.2 ... ) 2 ( ) 2 ( ) 2 ( 4 3 2 x x x forms a geometric series 1.2.1 Write down the common ratio (1) 1.2.2 Determine the value(s) of x for which the series will con
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Matched memo: 2025 G12 CT March Mpumalanga Memo.pdf
Relevant memo page(s): 3
[Page 3] n or n n n n n n n n n 17 T and 18 T are the terms equating factors 9 n answer (4) 2.2 486 , 162 , 54 3 2 1 T T T 52 a , 3 r , 7 n . 1 3 1 )3 ( 54 7 7 S 29538 series substitution 7 n answer (4) [14]
North West2025MarchTestMemo foundArithmeticQuadraticGeometric
2025 G12 CT March NW
Source folder: Tests
Source file: 2025 G12 CT March NW.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 2 2.1 Consider the quadratic pattern in diagram form below: 2.1.1 How many dots would you require to draw Picture 4? (1) 2.1.2 Determine the nth term of dots in pictures in the form Tn = an2 + bn + c. (4) 2.1.3 Between which two pictures will the difference between their numbers of dots be 74 ? (3) 2.2 Given the following arithmetic sequence: 483 ; ...... ; 11 ; 7 ; 3 . 2.2.1 Determine n T , the nth term of the sequence in the form: b an n T . (3) 2.2.2 How many terms are in the sequence. (2) 2.2.3 Determine the sum of the first 100 terms. (2) [Page 4] Mathematics/Test NSC - Grade 12 NW/March 2025 4 Please turn over 2.3 Given the geometric series: 3 2 2 1 1 1 k k x x Determine the value of x. (5) [20] QUESTION 3 3.1 Given: 1 1 2 ) ( x x f 3.1.1 Write down the equations of asymptotes of .f (2) 3.1.2 Sketch the graph of f , if points ) 0 ;1 ( and )1 ; 0 ( lie on f. Show all asymptotes. (3) 3.1.3 Write down the domain of 1 ) 2 ( x f (2) 3.1.4 Determine the equation of axis of symmetry of f for .0 m (2) 3.2 The exponential function is given by : 4 2 1 ) ( x x k 3.2.1 Is k increasing or decreasing. (1) 3.2.2 If k is translated 4 units up to form t(x), determine the equation of ) ( 1 x t , the inverse of k(x) in the form ... y (3) [13]
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Matched memo: 2025 G12 CT March NW Memo.pdf
Relevant memo page(s): 4
[Page 4] Mathematics/Test 1 MG NSC - Grade 12 NW/March 2025 4 QUESTION 2 2.1.1 29 answer (1) 2.1.2 1 3 1 3 1 3 1 5 6 )1( 3 2 2 2 n n T c b a c b a n value of a value of b value of c answer (4) 2.1.3 th 2 2 1 35 and 36 36 74 2 2 2 2 T 74 ....; 10; 8; ;6 or 35 and 36 36 72 2 74 2 2 74 ]1 )1 ( 3 )1 [( 1 3 74 th n th th n n n n n n n n n n n n T T 74 ]1 )1 ( 3 )1 [( 1 3 2 2 n n n n 36 n th 35 and 36th Or 74 ....; 10; 8; ;6 74 2 2 n th 35 and 36th (3) 2.2.1 1 4 ) 4 )( 1 ( 3 4 3 n n T d a n value of d correct substitution answer (3) 5 29 19 11 6 8 10 2 2
Northern Cape2025MarchTestMemo foundArithmeticQuadraticConvergence
2025 G12 NC March CT
Source folder: Tests
Source file: 2025 G12 NC March CT.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 2 2.1 Given the sequence 81 ; ; ;3 x y calculate the value(s) of x and y if : 2.1.1 The sequence is arithmetic. (4) 2.1.2 The sequence is geometric. (4) 2.2 The first four terms of a quadratic pattern are ...... 463 ; 525 ; 591 ; 661 2.2.1 Determine the nth term of this pattern. (4) 2.2.2 Calculate the value of the lowest term. (3) 2.3 An arithmetic sequence is given such that : 3029 ..... 2013 7 5 3 1 T T T T T and 6050 ..... 2014 8 6 4 2 T T T T T . Determine the common difference between each term. (3) [18] [Page 4] QUESTION 3 3.1 Prove that 0 23.7 k k converges. (3) 3.2 Hence , determine the value of m if 2 21 3.7 2 m k k (5) [8] QUESTION 4 Sketched below is the graph of 9 3 x x h for ) 2 ; 3 [ x 4.1 Write down the equation of the asymptote of h (1) 4.2 Determine the coordinates of A. (2) 4.3 Determine the equation of the straight line g, if g is passing through A and B. (3) 4.4 Write down the equation of 9 x h x m (1) 4.5 Write down the equation of x m 1 in the form y=...... (2) 4.6 Write down the range of x m 1 (2) 4.7 Point C is the result of point A when reflected over the line 2 x . Calculate the area ofΔABC. (4) [15]
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Matched memo: 2025 G12 NC March CT Memo.pdf
Relevant memo page(s): 4
[Page 4] 4 QUESTION 2 2.1.1 29 26 55 55 26 81 26 78 3 3 3 81 3 3 81 3; ; ; 81 y y x x d d d d a a y x 3 3 81 d 36 d 55 x 29 y (4) 2.1.2 9 3 1 81 3 1 27 27 3 1 81 3 1 27 1 3 81 3 81 3 ; ; ; 81 2 3 3 3 y OR y x x r r r ar a y x 3 81 3 r 3 1 r 27 x 9 y (4)
Western Cape2025MarchTestMemo foundArithmeticQuadraticGeometricConvergence
2025 G12 March Khayelitsha PLC
Source folder: Tests
Source file: 2025 G12 March Khayelitsha PLC.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 1 An arithmetic series has its 19th term equal to 11 and its 31st term is 5. 1.1 Determine the first term and the common difference. (4) 1.2 Hence, calculate the sum of the first 31 terms. (2) [6] QUESTION 2 A quadratic number pattern is defined by: m k n k P 2 ) 2 4 ( 2.1 Show that the first three terms of this quadratic number pattern are (3) 2.2 Determine the general term of this quadratic number pattern in the form . (4) 2.3 Determine the value of for which m k k 2 560 ) 2 4 ( . (5) [12] [Page 4] QUESTION 3 3.1 A geometric series is given as: ... Prove that the sum of the first terms is defined by 1 ; 1 )1 ( r r r a S n n (4) 3.2 Given a convergent geometric series ( ) ( ) ( ) ... 3.2.1 For which values of does the series converge? (3) 3.2.2 If 4 3 t , calculate . (3) [10] QUESTION 4 Prusent can do 25 push-ups per minute. Each week he improves his performance by 4 push-ups per minute. He intends to enter a Khayelitsha High Schools fitness competition that is in 10 weeks' time. 4.1 If he is consistent, how many push-ups will he do per minute in the fitness competition which is in 10 weeks' time? (2) 4.2 While training for the competition, Prusent decided to switch up and spent 45 minutes doing push-ups per week. If he now performs at this rate, how many push-ups does he do altogether in the 10 weeks? (
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Matched memo: 2025 G12 March Khayelitsha PLC Memo.pdf
Relevant memo page(s): 2
[Page 2] MATHEMATICS CT MEMO KHAYELITSHA PLC MARCH 2025 2 QUESTION 1 MARKS [6] 1.1 ( ) OR ( ) (4) (4) RP 1.2 [ ] [ ( )] OR [ ] [ ] Sub. into correct formula CA from 1.1 Answer (2) Sub. into correct formula CA from 1.1 Answer (2) K [6]
Free State2025JuneProvincial PaperEnglishMemo foundQuadraticGeometricSigma
2025 G12 Free State P1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 Free State P1 June.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 2 2.1 The th n term of the quadratic pattern is 2 2 nT n bn c = + + . The 2nd term of the quadratic pattern is 7. The 1st term of the first differences is 4. 2.1.1 Determine the first term of the quadratic pattern. (1) 2.1.2 Show that b = - 2 and c = 3. (2) 2.1.3 Which term of the pattern is equal to 4 903? (3) 2.1.4 Between which two consecutive terms of the quadratic pattern will the first difference be 2 000? (4) 2.1.5 Express the sum of the first 50 terms of the first differences in sigma notation. (2) 2.2 Determine the value of k if 4 (3 2) 913 k n n = - = . (6) [18] [Page 4] QUESTION 3 3.1 Prove that the sum to n terms of a geometric series, of which the first term is a and the common ratio r, can be given as: ( )1 1 n n a r S r - = - ; 1 r (3) 3.2 Given the geometric sequence: 9 27 2187 3 ; ; ;...; 2 4 64 3.2.1 Determine the nth term of the sequence. (2) 3.2.2 How many terms are in the sequence? (3) 3.3 The first three terms of a geometric series are: 2 3 4 8 4 2 ... x x x + + + 3.3.1 For which values of x will the series converge? (3) 3.3.2 If 72 7 S= , determine the possible value(s) of x. (4) 3.4 The sum of the first n terms of a sequence is 5 3 2 n n S - = + . Determine the value of the 80th term. Leave your answer in the form . c a b where a, b and c are integers. (4) [19]
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Matched memo: 2025 G12 Free State P1_V1 June Memo.pdf
Relevant memo page(s): 4
[Page 4] QUESTION 2 2.1.1 1 7 4 3 T = - = ✓ answer (1) 2.1.2 3(2) 4 b + = 2 b = - 2 2 3 c -+ = 3 c = ✓ substitution 2 a = ✓ correct substitution (2) 2.1.3 2 4903 2 2 3 n n = - + 2 0 2 2 4900 n n = - - 0 ( 50)( 49) n n = - + 50 n = or 49 n - ✓ equating ✓ factors ✓ answer with selection (3) 2.1.4 4 d = 2000 4 ( 1)4 n = + - 2000 4n = 500 n = 500 T & 501 T ✓ 4 d = ✓ substitution ✓ 500 n = ✓ answer (4) 2.1.5 50 1 4 k k = ✓4k ✓ 50 1 k = (2) 2.2 4 (3 2) 913 k n n = - = 4 1 3 n k k = -+ = - 3(4) 2 10 a = - = 3(5) 2 10 3 d = - - = 3 913 2(10) ( 3 1)(3) 2 k k - = - -- 1826 ( 3)[8 3 ] k k = - + 2 3 1850 0 k k - - = ( 25)(3 74) 0 k k - + = 74 25 or 3 k k = - ✓ n ✓ a ✓ d ✓ equating ✓ standard form ✓ answer with selection (6) [18]
Free State2025JuneProvincial PaperAfrikaansMemo foundGeometricConvergence
2025 G12 Free State V1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 Free State V1 June.pdf
Relevant question page(s): 4
[Page 4] VRAAG 3 3.1 Bewys dat die som tot n terme van 'n meetkundige reeks, waarvan die eerste term a en die gemeenskaplike verhouding r is, kan gegee word as: ( )1 1 n n a r S r - = - ; 1 r (3) 3.2 Gegee die meetkundige ry: 9 27 2187 3 ; ; ;...; 2 4 64 3.2.1 Bepaal die nde term van die ry. (2) 3.2.2 Hoeveel terme is in die ry? (3) 3.3 Die eerste drie terme van 'n meetkundige reeks is: 2 3 4 8 4 2 ... x x x + + + 3.3.1 Vir watter waardes van x sal die reeks konvergeer? (3) 3.3.2 Indien 72 7 S= , bepaal die moontlike waarde(s) van x. (4) 3.4 Die som van die eerste n terme van 'n reeks is 5 3 2 n n S - = + . Bepaal die waarde van die 80ste term. Los jou antwoord in die vorm . c a b waar a, b en c heelgetalle is. (4) [19]
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Matched memo: 2025 G12 Free State P1_V1 June Memo.pdf
Relevant memo page(s): 5
[Page 5] QUESTION 3 3.1 2 2 1 ...... n n n S a ar ar ar ar - - = + + + + + ...1 2 2 1 ) ... n n n n r rS ar ar ar ar ar - - = + + + + + ....2 n n n rS S ar a - = - ( ) ( ) 1 1 n n S r a r - = - ( )1 1 n n a r S r - = - ✓ n rS ✓ n n n rS S ar a - = - ✓ common factor (3) 3.2.1 1 3 3 2 n nT - = ✓ 3 2 r = ✓ 1 3 3 2 n- (2) 3.2.2 1 2187 3 3 64 2 n- = 1 729 3 64 2 n- = 6 1 3 3 2 2 n- = 7 n = ✓ equating ✓ simplification same base ✓answer (3) 3.3.1 1 2 r x = 1 1 1 2 x - 2 2 x - ✓ 1 2 r x = ✓ condition ✓ answer (3) 3.3.2 2 72 8 7 1 2 x x = - 2 56 72 36 x x = - 2 56 36 72 0 x x + - = 2 14 9 18 0 x x + - = (7 6)(2 3) 0 x x - + = 6 7 x = or 3 2 x = - ✓ subst. into correct formula ✓ standard form ✓factors ✓ x values (4)
Gauteng2025JuneProvincial PaperEnglishMemo foundArithmeticSigma
2025 G12 GP P1 June Second Push
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 GP P1 June Second Push.pdf
Relevant question page(s): 3, 4, 11
[Page 3] QUESTION 2 The first three terms of an arithmetic sequence are 4; 13 and 22. 2.1 Write down the fourth term of this sequence. (1) 2.2 Determine the general term of the sequence. (2) 2.3 Consider the terms of this sequence which are even. Calculate the sum of the first 25 terms which are even. (4) 2.4 The original sequence (4; 13 and 22) forms the first differences of a new sequence with a first term equal to - 6. Determine a formula for the nth term of this new sequence. (4) [11] ( ) ! " = - ! ! ! " # $ $ = - -! ! !"# ! # = ! ( )( ) ! " # > ! ! ( ) ! + = ! ! " ( ) ! " - - = ! ! " ( )! " ( ) ( )! " ! # = ( ) ( )! " ! # = [Page 4] QUESTION 3 3.1 Given: 3.1.1 Write down the values of the first three terms of the series. (2) 3.1.2 Write down the value of the constant ratio. (1) 3.1.3 Will converge? Explain your answer. (2) 3.1.4 Calculate . Give your answer in terms of x. (3) 3.2 and are the first three terms of an arithmetic sequence. 3.2.1 Determine the value of x. (5) 3.2.2 Calculate the value of the 10th term of this arithmetic sequence. (3) [16] ( ) ∑ = - !" # $ ! ! ( ) ∑ ∞ = - ! " ! ! ( ) ∑ = - !" # $ ! ! " ! " # ! - !" # + !
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Matched memo: 2025 G12 GP P2 June Second Push Memo.pdf
Relevant memo page(s): 3
[Page 3] 3 1.5 = - = = 4, 153 T 2 1 T tan 2 PT 2 m Equation of NR: y = 2 O Tˆ N S Nˆ R = (alt s; NR || OT) = S Nˆ R O Tˆ N = 180° - 153,4° = 26,6° PT 2 T tan m = 2 T = 4, 153 y = 2 S = 26,6 S Nˆ R (5) [16] QUESTION 2 2.1 A(-3 ; 4) x = -3 y = 4 (2) 2.2 r2 = (3)2 + (-4)2 r2 = 25 Equation of the circle through A, B and C: x2 + y2 = 25 substitution/ r2 = 25 answer (3) 2.3 r = 5 AB = 10 units r = 5 answer (2) 2.4 AB ⊥ ED (radius ⊥ tangent) eenhede units/ 5 BD 25 BD ) 10 ( ) 125 ( BD AB AD BD 2 2 2 2 2 2 2 = = - = - = S/R subst into Pyth th answer (3) 2.5 area of ∆ABD = hoogte basis height/ base/ 2 1 ⊥ eenhede vk units/ square 25 (5)(10) 2 1 = = formula substitution answer (3) 2.6 AE is a diameter (converse s in semi circle) AE= 4 - (-8,5) = 12,5 units r = 4 25 or/ 4 1 6 of Centre of new circlel: = ) 2 8,5 4 ; 3 ( - + - = )
Gauteng2025JuneProvincial PaperEnglishMemo foundQuadraticGeometricConvergence
2025 G12 GP P1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 GP P1 June.pdf
Relevant question page(s): 3, 4
[Page 3] next two terms of the pattern. (2) 2.1.2 Show that the sum of the first n terms of the pattern is given by: S= 1 2 n(7n -17) (3) 2.2 The first four terms of a quadratic pattern are: x; 3x-5 ; 4x-3 ; 5x+ 1 ; ... 2.2.1 Determine the value of x. (3) 2.2.2 If the pattern continues indefinitely, prove that all the terms of the pattern are positive. (5) [Page 4] MATHEMATICS (PAPER 1) GR12 0625 4 Please turn over 2.3 Consider the geometric series: 1 2 (p-3) + 3 4 (p-3)ଶ+ 9 8 (p-3)ଷ+ ⋯; for p≠3 2.3.1 Determine the values of pfor which the series converges. (4) 2.3.2 If the sum to infinity of the series is 1, determine the value of p. (3) [20] QUESTION 3 3.1 If 2(3ିଵ) = 59 046, ୀଶ determine the value of n. (5) 3.2 An equilateral triangle RST with sides of length 12p units is drawn. A second triangle is drawn by joining the midpoints of the sides of the first triangle RST. Each triangle thereafter is drawn by joining the midpoints of the sides of the previous triangle as shown on the sketch, and this continues indefinitely. 3.2.1 Write down, in terms of p, the length of each side of the second triangle. (1) 3.2.2 Calculate, in terms of p, the perpendicular height of
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Matched memo: 2025 G12 GP P1 June Memo.pdf
Relevant memo page(s): 10
[Page 10] QUESTION 3 3.1 ∑2(3k-1) = 59 046 . n k=2 6 + 18 + 54 + ⋯+ 2(3n-1) = 59 046 r= 18 6 = 3 Number of terms (n-2) + 1 = n-1 Sn= a(rn-1) r-1 ; r≠1 Sn-1 = 6(3n-1 -1) 3 -1 6(3n-1 -1) 3 -1 = 59 046 3(3n-1 -1) = 59 046 3n-1 -1 = 19 682 3n-1 = 19 683 3n-1 = 39 n-1 = 9 ∴n= 10 OR ∑2(3k-1) = 59 046 . n k=2 6 + 18 + 54 + ⋯+ 2(3n-1) = 59 046 r= 18 6 = 3 Number of terms (n-2) + 1 = n-1 Let the number of terms be k Sn= a(rn-1) r-1 ; r≠1 Sk= 6(3k-1) 3 -1 6(3k-1) 3 -1 = 59 046 3(3k-1) = 59 046 3k-1 = 19 682 3k= 19 683 3k= 39 k= 9 ∴n-1 = 9 ⇒n= 10 r= 3 n-1 substitution simplification to 3n-1 = 19 683 10 OR r= 3 n-1 substitution simplification to 3k= 19 683 10 If k= 9 max 4 5 (5)
Gauteng2025JuneProvincial PaperAfrikaansMemo foundGeometricConvergence
2025 G12 GP V1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 GP V1 June.pdf
Relevant question page(s): 4
[Page 4] WISKUNDE (VRAESTEL 1) GR12 0625 4 Blaai om asseblief 2.3 Beskou die meetkundige reeks: 1 2 (p-3) + 3 4 (p-3)ଶ+ 9 8 (p-3)ଷ+ ⋯; vir p≠3 2.3.1 Bepaal die waardes van pwaarvoor die reeks konvergeer. (4) 2.3.2 As die som tot oneindig van die reeks 1 is, bepaal die waarde van p. (3) [20] VRAAG 3 3.1 As 2(3ିଵ) = 59 046 , ୀଶ bepaal die waarde van n. (5) 3.2 'n Gelyksydige driehoek RST met sye van lengte 12peenhede word getoon. 'n Tweede driehoek word geteken deur die middelpunte van die sye van die eerste driehoek RST te verbind. Elke driehoek word daarna geteken deur die middelpunte van die sye van die vorige driehoek te verbind, soos op die skets getoon. Hierdie patroon gaan ongeindig voort. 3.2.1 Skryf, in terme van p, die lengte van elke sy van die tweede driehoek neer. (1) 3.2.2 Bereken, in terme van p, die
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Matched memo: 2025 G12 GP V1 Memo.pdf
Relevant memo page(s): 10
[Page 10] VRAAG 3 3.1 ∑2(3k-1) = 59 046 . n k=2 6 + 18 + 54 + ⋯+ 2(3n-1) = 59 046 r= 18 6 = 3 Aantal terme(n-2) + 1 = n-1 Sn= a(rn-1) r-1 ; r≠1 Sn-1 = 6(3n-1 -1) 3 -1 6(3n-1 -1) 3 -1 = 59 046 3(3n-1 -1) = 59 046 3n-1 -1 = 19 682 3n-1 = 19 683 3n-1 = 39 n-1 = 9 ∴n= 10 OF ∑2(3k-1) = 59 046 . n k=2 6 + 18 + 54 + ⋯+ 2(3n-1) = 59 046 r= 18 6 = 3 Aantal terme(n-2) + 1 = n-1 Stel k die aantal terme Sn= a(rn-1) r-1 ; r≠1 Sk= 6(3k-1) 3 -1 6(3k-1) 3 -1 = 59 046 3(3k-1) = 59 046 3k-1 = 19 682 3k= 19 683 3k= 39 k= 9 ∴n-1 = 9 ⇒n= 10 r= 3 n-1 substitusie vereenvoudiging tot 3n-1 = 19 683 10 Of r= 3 n-1 substitusie vereenvoudiging tot 3k= 19 683 10 (5)
KwaZulu-Natal2025JuneProvincial PaperEnglishMemo foundArithmeticQuadratic
2025 G12 KZN P1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 KZN P1 June.pdf
Relevant question page(s): 3
[Page 3] QUESTION 2 2.1 Given the arithmetic series: 6 + 1 4 9 . . . . . . . 2.1.1 Write down the value of the next term of the arithmetic series. (1) 2.1.2 Calculate: 6 + 1 4 9 . . . . . . . . . 239 (5) 2.2 Consider a quadratic pattern: 9 ; 5 ; ; 15 ; ... x 2.2.1 Calculate the value of x. (3) 2.2.2 If the value of 3 x , determine the nth term of the number pattern. (4) 2.2.3 Explain why all the terms of this quadratic pattern are odd numbers. (2) [15]
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Matched memo: 2025 G12 KZN P1 June Memo.pdf
Relevant memo page(s): 4
[Page 4] Mathematics P1 June 2025 Examination GRADE 12 Marking Guidelines Copyright Reserved Please turn over 4 2 2 2 0 x px p 2 2 2 2 4 2 4 2 2 2 9 4 3 4 2 b b ac x a p p p p p p p p or p The roots are rational A substitution into quadratic formula CA 2 9 4 p p CA 3 4 p p (3) [26] QUESTION 2 2.1.1 5 14 T A answer (1) 2.1.2 1 nT a n d 6 1 5 239 n 6 5 5 239 n 5 250 n 50 n 2 n n S a l OR 2 1 2 n n S a n d 50 50 6 239 2 S 50 50 2 6 50 1 5 2 S 5825 50 25 233 S 5825 A substitution in nT CA equating nT to 239 CA value of n CA substitution in nS CA answer (5) 2.2.1 5 4 15 5 3 9 3 x x x x x A 1st differences: 4; x + 5; 15 - x CA 5 4 15 5 x x x CA answer (3) - 9 - 5 x 15 4 x + 5 15 - x
KwaZulu-Natal2025JuneProvincial PaperAfrikaansMemo foundArithmetic
2025 G12 KZN V1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 KZN V1 June.pdf
Relevant question page(s): 3
[Page 3] VRAAG 2 2.1 Die rekenkundige reeks word gegee: 6 + 1 4 9 . . . . . . . - - 2.1.1 Skryf die waarde van die volgende term van die rekenkundige reeks neer. (1) 2.1.2 Bereken: 6 + 1 4 9 . . . . . . . . . 239 - - - (5) 2.2 Beskou 'n kwadratiese patroon: 9 ; 5 ; ; 15 ; ... x - - 2.2.1 Bereken die waarde van x. (3) 2.2.2 Indien die waarde van 3 x = , bepaal die nde term van die getalpatroon. (4) 2.2.3 Verduidelik waarom al die terme van dié kwadratiese patroon onewe getalle is. (2) [15]
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Matched memo: 2025 G12 KZN P1 June Memo.pdf
Relevant memo page(s): 4
[Page 4] Mathematics P1 June 2025 Examination GRADE 12 Marking Guidelines Copyright Reserved Please turn over 4 2 2 2 0 x px p 2 2 2 2 4 2 4 2 2 2 9 4 3 4 2 b b ac x a p p p p p p p p or p The roots are rational A substitution into quadratic formula CA 2 9 4 p p CA 3 4 p p (3) [26] QUESTION 2 2.1.1 5 14 T A answer (1) 2.1.2 1 nT a n d 6 1 5 239 n 6 5 5 239 n 5 250 n 50 n 2 n n S a l OR 2 1 2 n n S a n d 50 50 6 239 2 S 50 50 2 6 50 1 5 2 S 5825 50 25 233 S 5825 A substitution in nT CA equating nT to 239 CA value of n CA substitution in nS CA answer (5) 2.2.1 5 4 15 5 3 9 3 x x x x x A 1st differences: 4; x + 5; 15 - x CA 5 4 15 5 x x x CA answer (3) - 9 - 5 x 15 4 x + 5 15 - x
Limpopo2025JuneProvincial PaperEnglishMemo foundArithmeticQuadratic
2025 G12 Limpopo P1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 Limpopo P1 June.pdf
Relevant question page(s): 4
[Page 4] Mathematics/P 1 QUESTION2 4 NSC LimpopoDoE/June 2025 2-1 The 19th term of an arithmetic sequence is 11, while the 31 st term is 5 • 2.1.1 2.1.2 2.1.3 Determine the first 3 terms of the sequence. Determine the general term of the sequence. Calculate the sum of the first 81 terms. 2.2 Given the quadratic pattern: 2 ; 5 ; 10 ; 17 ; 26 ; ........ .. 2.2.1 Write down the next TWO terms of the pattern. (5) (2) (3) (2) 2.2.2 Show that ~ = n2 + 1 is the general term of the quadratic pattern. (4) 2.2.3 2.2.4 Determine the term of the pattern that has the value of 290. Between which two consecutive terms of the quadratic pattern will the first difference be 25 ? QUESTION3 Square A has sides 2cm each. The length of the side of square A is half the length of the side of square B, and the length of the side of square C is
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Matched memo: 2025 G12 Limpopo P1 June Memo.pdf
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[Page 6] reenvoudig answer/antwoord (3) 2.2 2.2.1 37 and / 50 en 37 50 (2) 2.2.2 2 5 10 17 26 3 5 7 9 2 2 2 2 2 2 1 3 3 3(1) 3 0 2 1 0 2 1 1 n a a a b b b a b c c c T n second difference/tweede verskil 2 2 a 3 3 a b 2 a b c (4)
Limpopo2025May / JuneProvincial PaperEnglishMemo foundPatterns & Sequences
2025 G12 Limpopo P1 Pre June Exam
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 Limpopo P1 Pre June Exam.pdf
Relevant question page(s): Not isolated from scan
Topic section could not be extracted cleanly from this PDF scan, but this source paper is still included from the local Grade 12 resource bank.
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Matched memo: 2025 G12 Limpopo P1 June Memo.pdf
Relevant memo page(s): 6
[Page 6] reenvoudig answer/antwoord (3) 2.2 2.2.1 37 and / 50 en 37 50 (2) 2.2.2 2 5 10 17 26 3 5 7 9 2 2 2 2 2 2 1 3 3 3(1) 3 0 2 1 0 2 1 1 n a a a b b b a b c c c T n second difference/tweede verskil 2 2 a 3 3 a b 2 a b c (4)
Mpumalanga2025JuneProvincial PaperEnglishMemo foundArithmeticGeometric
2025 G12 Mpumalanga P1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 Mpumalanga P1 June.pdf
Relevant question page(s): 4
[Page 4] 12,5 km. After 7 weeks he can no longer increase the distance he cycles. What is the maximum distance Wiggil can cycle per (3) week? 2.2 Consider the following sequence: 4; 10; 18; 28; ... 2.2.1 Determine a formula for the nth term of the sequence. 2•2•2 What will be the value of the 17th term ? QUESTION 3 3 .1 x; 4x + 5; 1 Ox - 5 are the first three terms of an arithmetic sequence. Determine the value of x and hence the numeric sequence. 3.2 Given the arithmetic series 2 + 6 + 10 + ... + 398 Calculate: 3 .2.1 The number of terms in this series. 3.2.2 The sum of the 30 terms in the middle of this series. 3.3 Calculate the value of a if: 3 Z:a.2k-l =28 k=I 3.4 If~ = i-n, calculate S00 • (4) (2) (9) (4) (3) (4) (2) (4) 3.5 Given that Sn =7; + 1'z + .......... In-1 + ~ is a geometric series. Show that the formul
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Matched memo: 2025 G12 Mpumalanga P1_V1 June Memo.pdf
Relevant memo page(s): 4
[Page 4] VRAAG 2 2.1 ..... .......... 65 ; 5, 52 ; 40 + )1 (5, 12 40 - + = n Tn 7 40 12,5(7 1) T = + - 115 = km ✓sequence / ry ✓substitution of a and d / vervang a en d ✓answer / antwoord (3) 2.2.1 2 2 = a 6 3 = +b a 4 = + + c b a 1 = a 6 )1(3 = +b 4 3 1 = + + c 3 = b 0 = c n n Tn 3 2 + = ✓ 1 = a ✓ 3 = b ✓ 0 = c ✓ n n Tn 3 2 + = (4) 2.2.2 ) 17 (3 ) 17 ( 2 17 + = T 340 = ✓ substitution / vervanging ✓ answer / antwoord (2) [9] QUESTION 3/VRAAG 3 3.1 5 10 ; 5 4 ; - + x x x x x - +5 4 ) 5 4 ( 5 10 + - - x x = - + x x 5 4 ) 5 4 ( 5 10 + - - x x 5 = x Sequence/Ry: . 45 ; 25 ; 5 ✓difference / verskil ✓equating / vergelyking ✓ 5 = x ✓sequence / ry (4) 3.2.1 2 6 10 ............. 398 + + + + )1 ( - + = n d a Tn )1 ( 4 2 - + = n 398 2 4 4 n = + - 4 400 n = 100 n = ✓substitution of d / vervanging van d 398 nT = ✓answer / antwoord
Mpumalanga2025JuneProvincial PaperAfrikaansMemo foundPatterns & Sequences
2025 G12 Mpumalanga V1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 Mpumalanga V1 June.pdf
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A matching memo file was found, but the answer section could not be isolated cleanly from the current PDF scan.
North West2025JuneProvincial PaperEnglishMemo foundQuadraticGeometric
2025 G12 NW P1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 NW P1 June.pdf
Relevant question page(s): 3, 4
[Page 3] VRAAG 2 Gegee is die kwadratiese getalpatroon: 2; 7; 16; . . . . . . ; 862 2.1 Skryf die 4de term van die kwadratiese getalpatroon neer. (1) 2.2 Bepaal die algemene term ) (Tn vir die kwadratiese getalpatroon. (4) 2.3 Bereken die aantal terme in die getalpatroon. (3) [8] [Page 4] VRAAG 3 3.1 Gegee die meetkundige reeks: 2 + 6 + 18 + 54 + . . . 3.1.1 Skryf die algemene term van hierdie reeks neer. (1) 3.1.2 Bereken die waarde van m sodanig dat: m n n 1 048 59 3 3 2 (4) 3.2 Die eerste term van 'n oneindige meetkundige ry is 3 en die gemeenskaplike verhouding van die ry is 1 2. 3.2.1 Bepaal die waarde van die derde term van die ry. (1) 3.2.2 Bepaal die waarde van S∞. (2) 3.3 Indien die magte van 6 uit die ry verwyder word 1; 2; 3; 4; . . . ; 8 000, bepaal die som van die oorblywende terme. (4) [12]
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Matched memo: 2025 G12 NW P1_V1 June Memo.pdf
Relevant memo page(s): 3
[Page 3] Mathematics P1/Wiskunde V1 3 NW/June/Junie 2025 Grade/Graad 12 - Marking Guidelines/Nasienriglyne Copyright reserved Please turn over 1.1.4 6 -√x+ 4 = x+ 4 -√x+ 4 = x-2 x+ 4 = x2 -4x+ 4 x2 -5x= 0 x(x-5) = 0 x= 0 or/of x≠5 isolate/isoleer √ square/kwadreer std form/vorm factors/faktore x= 0 or/of x= 5 x≠5 (6) 1.2 2y-x= 3 . . . . . . . . . 1 y2 + 3x-2xy= 0 . . . . 2 2y-3 = x . . . . . . . . . 3 ∴y2 + 3(2y-3) -2y(2y-3) = 0 y2 + 6y-9 -4y2 + 6y= 0 -3y2 + 12y-9 = 0 y2 -4y+ 3 = 0 (y-1)(y-3) = 0 y= 1 or/of y= 3 x= 2(1) -3 or/of x= 2(3) -3 = -1 = 3 equation/vergelyking subst std form/vorm factors/faktore y-values/waardes x-values/waardes (6) 1.3 (3 -1 √5 ) (9 + 3 √5 + 1 5) = 33 -( 1 √5 ) 3 = 27 -1 5 ∙1 √5 × √5 √5 = 27 -1 25 √5 ∴a= 27 and/en b= 5 cubes/derdemagte rationalise/rasionaliseer a= 27 b= 5 (4) [27] QUESTION/V
North West2025JuneProvincial PaperEnglishMemo foundArithmeticQuadraticGeometric
2025 G12 NW Rustenburg Leo P1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 NW Rustenburg Leo P1 June.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 2 2.1 The first FOUR terms of a quadratic pattern are: 15 ; 29 ; 41 ; 51 2.1.1 Write down the value of the 5th term. (1) 2.1.2 Determine an expression for the nth term of the pattern in the form c bn an Tn + + = 2 . (4) 2.1.3 Determine the value of 27 T (2) a b c [Page 4] Mathematics/ P1 NSC RTB LEO/May 2025 4 2.2 Given a geometric sequence: 36 ; -18 ; 9 ; ... 2.2.1 Determine the value of r, the common ratio. (1) 2.2.2 Calculate n if 096 4 9 T = n (3) 2.2.3 Calculate S (2) 2.2.4 Calculate the value of 500 8 6 4 2 499 7 5 3 1 T ... T T T T T ... T T T T + + + + + + + + + + (4) [17] QUESTION 3 3.1 The first three terms of an arithmetic sequence are: 2p + 3 ; p + 6 and p - 2. 3.1.1 Show that p = 11. (2) 3.1.2 Calculate the smallest value of n for which . 55 - n T (3) 3.2 Given that = = - = - 9 1 6 1 ) 3 ( ) 3 ( k k k x k x , prove that = = - 15 1 0 ) 3 ( k k x . (5) [10] QUESTION 4 Given the exponential function: x x g = 2 1 ) ( 4.1 Write down the range of g. (1) 4.2 Determine the equation of 1 - g in the form y = ... (2) 4.3 Is 1 - g a function? Justify your answer. (2) 4.4 The
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Matched memo: 2025 G12 NW Rustenburg Leo P1 June Memo (Unofficial).pdf
Relevant memo page(s): Not isolated
A matching memo file was found, but the answer section could not be isolated cleanly from the current PDF scan.
Northern Cape2025JuneProvincial PaperEnglishMemo foundQuadraticGeometricConvergence
2025 G12 NC P1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 NC P1 June.pdf
Relevant question page(s): 4
[Page 4] QUESTION 2 2.1 Consider the following quadratic pattern: x ; 17 ; y ; 57 ; 86. Determine the values of x and y. (5) 2.2 An athlete trains by running 600 m on the first day. Thereafter, she increases the distance by 300 m every day. (This is a theoretic scenario.) 2.2.1 Calculate the distance she runs on the 15th day. (3) 2.2.2 What is the total distance, in km, that she has run at the end of the 15th day? (3) 2.2.3 In order to participate in the Comrades Marathon in 6 months (180 days), she must complete a qualifying race of 42 km. If she continues her pattern of training, will she have sufficient time to train for the 42 km qualifier and thus be able to participate in the Comrades? Show ALL calculations. (3) 2.3 Use an appropriate formula to calculate: 21 1 (101 5 ) n n (3) 2.4 Given: A convergent geometric
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Matched memo: 2025 G12 NC P1_V1 June Memo.pdf
Relevant memo page(s): 5
[Page 5] Mathematics P1/Grade 12/Wiskunde V1/Graad 12 5 NC/June 2025/NK/Junie 2025 Marking Guidelines/Nasienriglyne Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief QUESTION/VRAAG 2 2.1 x 17 y 57 86 17 - x y - 17 57 - y 29 y - 34 + x 74 - 2y - 28 + y 74 2 28 3 102 34 34 74 2 34 34 74 2(34) 6 y y y y y x y x x - = - + = = - + = - - + = - = ✓setting 1st diffences/stel van 1ste verskille ✓setting 2nd diffences/stel van 2de verskille ✓equating/vergelyk ✓answer/antwoord ✓answer/antwoord (5) 2.2.1 600 900 1200 300 300 15 600 (14)(300) 4800 4,8 T m OR km = + = = OR 15 300 300 300(15) 300 = 4800 m nT n T = + = + ✓finding ''d''/ vind ''d'' substitution into the correct formula/substitusie in die korrekte formule answer/antwoord (3) 2.2.2 15 15[2(600) (14)(300)] 2 40 500 40,5 S m km = + = = substitution
Northern Cape2025JuneProvincial PaperAfrikaansMemo foundGeometric
2025 G12 NC V1 June
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 NC V1 June.pdf
Relevant question page(s): 4
[Page 4] p en dus aan die Comrades kan deelneem? Toon ALLE bewerkings. (3) 2.3 Gebruik 'n geskikte formule en bereken: 21 1 (101 5 ) n n (3) 2.4 Gegee: 'n Konvergerende meetkundige reeks met 'n eerste term 1T a en S p waar p > 0. 2.4.1 Toon dat (0 ; 2 ) a p . (5) 2.4.2 Bepaal die waarde van die konstante verhouding indien 4 p a . (3) [25]
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Matched memo: 2025 G12 NC P1_V1 June Memo.pdf
Relevant memo page(s): Not isolated
A matching memo file was found, but the answer section could not be isolated cleanly from the current PDF scan.
SACAI2025May / JuneSACAI PaperEnglishMemo foundQuadraticGeometricConvergence
2025 G12 SACAI P1 May_June
Source folder: 2024 - 2025 SACAI Exams
Source file: 2025 G12 SACAI P1 May_June.pdf
Relevant question page(s): 4
[Page 4] QUESTION 2 2.1 The 4th term of a geometric sequence is 6, and the 9th term is 0,1875. Determine the sequence. (5) 2.2 Calculate the sum of all the integers from 100 to 300 that are multiples of 4. (5) 2.3 The first two terms of a converging geometric series are 8 and m. The sum to infinity of the series is 12. Determine the constant ratio of the series. (6) [16] QUESTION 3 Given the quadratic sequence: -12; -11; -8; -3 ... 3.1 Determine the next two terms in the sequence. (3) 3.2 Determine the general term of the quadratic sequence. (4) 3.3 Determine the general term of the first differences of the sequence. (2) 3.4 Determine the difference between T60 and T61 of the quadratic sequence. (3) [12]
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Matched memo: 2025 G12 SACAI P1 May_June Memo.pdf
Relevant memo page(s): 3
[Page 3] QUESTION 2 2.1 T9 = ar8 = 0,1875 T4 = ar3 = 6 ∴r5 = 1 32 r= 1 2 and a= 48 48; 24; 12; ... ✔ T9 = ar8 ✔ T4 = ar3 ✔ r5= 1 32 ✔ a and r ✔ Sequence (5) 2.2 100 + 104 + 108 + ⋯+ 300 300 = 100 + (n-1)4 4n= 204; n= 51 S51 = 51 2 (100 + 300) = 10200 OR S51 = 51 2 (2(100) + 50(4) = 10200 ✔ Sequence ✔ Subst in formula ✔ n= 51 ✔ Subst in formula ✔ Answer (5) 2.3 r= m 8 S∞= 8 1-m 8 = 12 8 = 12(1 - m 8) 8 = 12 - 3m 2 8 = 3m m= 8 3 r= 8 3 × 1 8 = 1 3 ✔ r= m 8 ✔Correct formula with correct substitution ✔ = 12 ✔ Simplify ✔ m ✔Answer (6) [16]
SACAI2025May / JuneSACAI PaperAfrikaansMemo foundQuadraticGeometric
2025 G12 SACAI V1 Mei_Junie
Source folder: 2024 - 2025 SACAI Exams
Source file: 2025 G12 SACAI V1 Mei_Junie.pdf
Relevant question page(s): 4
[Page 4] VRAAG 2 2.1 Die 4de term van ʼn meetkundige ry is 6, en die 9de term is 0,1875. Bepaal die ry. (5) 2.2 Bereken die som van al die heelgetalle vanaf 100 tot 300 wat veelvoude van 4 is. (5) 2.3 Die eerste twee terme van ʼn konvergerende meetkundige reeks is 8 en m. Die som tot oneindig van die reeks is 12. Bepaal die konstante verhouding van die reeks. (6) [16] VRAAG 3 Gegee die kwadratiese ry: -12; -11; -8; -3 ... 3.1 Bepaal die volgende twee terme van die ry. (3) 3.2 Bepaal die algemene term van die kwadratiese ry. (4) 3.3 Bepaal die algemene term van die eerste verskille van die ry. (2) 3.4 Bepaal die verskil tussen T60 en T61 van die kwadratiese ry. (3) [12]
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Matched memo: 2025 G12 SACAI V1 Mei_Junie Memo.pdf
Relevant memo page(s): 3
[Page 3] NSS MEI/JUNIE: WISKUNDE V1 NR| 2025 © SACAI KOPIEREG Bladsy 3 van 11 Blaai asb. om VRAAG 2 2.1 T9 = ar8 = 0,1875 T4 = ar3 = 6 ∴r5 = 1 32 r= 1 2 en a= 48 48; 24; 12; ... ✔ T9 = ar8 ✔ T4 = ar3 ✔ r5= 1 32 ✔ a en r ✔ Ry (5) 2.2 100 + 104 + 108 + ⋯+ 300 300 = 100 + (n-1)4 4n= 204; n= 51 S51 = 51 2 (100 + 300) = 10200 OF S51 = 51 2 (2(100) + 50(4) = 10200 ✔ Ry ✔ Vervang in formule ✔ n= 51 ✔ Vervang in formule ✔ Antwoord (5) 2.3 r= m 8 S∞= 8 1-m 8 = 12 8 = 12(1 - m 8) 8 = 12 - 3m 2 8 = 3m m= 8 3 r= 8 3 × 1 8 = 1 3 ✔ r= m 8 ✔ Korrek formule met korrekte vervanging ✔ = 12 ✔ Vereenvoudig ✔ m ✔ Antwoord (6) [16]
SACAI2025AugustSACAI PaperAfrikaansMemo not linkedQuadraticGeometricSigma
2025 G12 SACAI V1 Aug
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 SACAI V1 Aug.pdf
Relevant question page(s): 4, 5, 14
[Page 4] VRAAG 2 Die eerste vier terme van 'n kwadratiese getalpatroon is: 94; x; 82; x-20 2.1 Toon deur berekeninge aan dat x= 90. (4) 2.2 Wys dat die nde term van die patroon geskryf kan word as: Tn= -2 (n- 1 2) 2 + 189 2 (6) 2.3 Bepaal die waarde van T15. (2) 2.4 Die EERSTE verskille vorm 'n rekeningkundige ry. Bepaal watter twee opeenvolgende terme van die patroon 'n EERSTE verskil van -40 sal hê. (4) [16] [Page 5] VRAAG 3 3.1 Die volgende meetkundige ry word gegee: 12; 4; 4 3 ; 4 9 ; ... ; 4 59 049 3.1.1 Bepaal die waarde van r, die gemene verhouding. (1) 3.1.2 Hoeveel terme is daar in die ry? Toon alle berekeninge. (3) 3.1.3 Bepaal S∞. (2) 3.2 Die 3de term van 'n meetkundige reeks is 64 en die 5de term is 256. Bepaal die som van die eerste 8 terme, waar r> 0. (4) [10]
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No confident matching memo PDF was found for this source file inside the current local non-investigation resource bank.
Limpopo2025Pre-TrialProvincial PaperEnglishMemo foundQuadraticConvergence
2025 G12 Limpopo P1 Pre-Trial
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 Limpopo P1 Pre-Trial.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 2 2.1. Given the sequence: 16 ; 22 ; 28 ;........ 2.1.1 Determine the formula for the th n term of the sequence. (3) 2.1.2 Calculate the 100th term of the sequence. (2) 2.1.3 Determine the term of the sequence which is equal to 310. (3) 2.2 The first three terms of a quadratic pattern are: 8 ; ; 4 ; 24 x x 2.2.1 Calculate the value(s) of x. (3) 2.2.2 If 5, x determine the position of the first term in the quadratic number pattern for which the sum of the first th n term difference will be greater than 300 . (4) [15] + [Page 4] QUESTION 3 Given: 1 1 4(0,2 )h h x 3.1 Write the first three terms of the series. (1) 3.2 For which values of x will the series converge? (2) 3.3 Calculate the S in terms of x. (3) 3.4 If 1, x calculate the smallest number of terms of the series whose sum will differ by less than 0,0001 from the sum to infinity of the series. (5) [11] QUESTION 4 The graph of 2 ( ) 5 and ( ) 2 15 f x x g x x x are drawn below. A and B are the x-intercepts and D is the y-intercept of g. E is the turning point of g and K is the y-intercept of f . Line EL is to the x-axis. 4.1 Determine: 4.1.1 The co-ordinates of A, B and K. (3) 4.1.2 The co-ordinates of D (1) 4.1.3 The co-ordinates of E (3) 4.2 Determine the range of f . (1) 4.3 For which value(s) of x is ( ) 0 g x (2) 4.4 Determine the coordinates of C, the point of intersectio
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Matched memo: 2025 G12 Limpopo P1 Pre-Trial Memo.pdf
Relevant memo page(s): 5
[Page 5] 50 nT n n n equating/gelykstel substitution/vervanging answer/antwoord (3) 2.2 2.2.1 8 4 24 x x 8 4 20 x x x 4 2 2 16 x x 2 4 2 16 2 2 16 4 4 20 5 x x x x x x second difference/2e verskil equating/gelykstel 5 x (3)
Mpumalanga2025Pre-TrialProvincial PaperEnglishMemo foundArithmeticQuadraticSigma
2025 G12 Mpumalanga (Ehlanzeni District) P1 Pre-Trial
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 Mpumalanga (Ehlanzeni District) P1 Pre-Trial.pdf
Relevant question page(s): 3, 4, 10
[Page 3] QUESTION 2 2.1 Prove that for any arithmetic sequence of which the first term is a and the constant difference is d, the sum to n terms can be expresses as: Sn= n 2 [2a+ (n-1)d] (4) 2.2 Consider the sequence: 1 2 ; 4 ; 1 4 ; 7 ; 1 8 ; 10; ... 2.2.1 If the pattern continues in the same way, write down the next TWO terms in the sequence. (2) 2.2.2 Calculate the sum of the first 50 terms of the sequence. (7) 2.3 Determine the value(s) of x, for which the series: 3 + 3(x+ 1) + 3(x+ 1)2 + ⋯ converge. (3) [16] [Page 4] QUESTION 3 3.1 A quadratic number pattern Tn= an2 + bn+ c has a first term equal to 1. The general term of the first differences is given by 4n+ 6. 3.1.1 Determine the value of a. (2) 3.1.3 Determine the formula for Tn. (4) 3.2 Given the series: (1 × 2) + (5 × 6) + (9 × 10) + (13 × 14)+ ... + (81 × 82) Write the series in sigma notation. (It is not necessary to calculate the value of the series). (4) [10] QUESTION 4 The lines y= x+ 1 and y= -x-7 are the axes of symmetry of the function f(x) = -2 x+p+ q. 4.1 Show that p= 4 and q= -3. (4) 4.2 Calculate the x-intercept of f. (2) 4.3 Sketch the graph of f. Clearly label ALL intercept with the axes and asymptotes. (3) [9]
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Matched memo: 2025 G12 Mpumalanga (Ehlanzeni District) P1 Pre-Trial Memo.pdf
Relevant memo page(s): 4
[Page 4] Mathematics P1/Memorandum 3 Ehlanzeni/August 2025 FET-GRADE 12 Copyright reserved Please turn over 1.3 Let the shortest side be x Dimension of the box: l= x; b= 2x ; h= 3x V = lbh 3 072 = (x)2x)(3x) 6x3 = 3 072 x3 = 512 x= 8 shortest side be x x; 2x; 3x substitute in volume formula answer (4) [24] QUESTION 2 2.1 Sn= a+ [a+ d] + [a+ 2d] + ⋯+ [a+ (n-2)d+ [a+ (n-1)d Sn= [a+ (n-1)d ]+[ a+ (n-2)d]+...+[a+ d] + a 2Sn= [2a+ (n-1)d]+[2a+ (n-1)d] 2Sn = n[2a+ (n-1)d] Sn= n 2 [2a+ (n-1)d] Sn Reversing Sn Expressing 2Sn Grouping to get 2Sn= n[2a+ (n-1)d] (4) 2.2.1 1 16 ; 13 answers (2) 2.2.2 (1 2 + 1 4 + 1 8 + ⋯to 25 terms) (4 + 7 + 10 + ⋯to 25 terms) Sn= a(rn-1) r-1 Sn= n 2 [2a+ (n-1)d] S25 = 1 2((1 2) 25 -1) 1 2-1 S25 = 25 2 [2(4) + (25 -1)3] S25 = 0,999999 S25 = 1 000 S50 = 0,999999 + 1000 S50 = 1001,00 OR a & r Geome
Eastern Cape2025SeptemberProvincial PaperEnglishMemo foundQuadraticGeometricConvergence
2025 G12 EC P1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 EC P1 Sept.pdf
Relevant question page(s): 4, 5
[Page 4] QUESTION 2 O lcarmers, Amanda and Leroy were given a task to come up with any common ratio (r) that would be suitable lo calculate the sum to infinity ifit given that the first term, a 2 Below are the calculations Amanda's calculation: =2 and r=3 2.1.1 2.1.2 S Leroy's calculation: a=2 and r= S 2.2 Evaluate: 2 Copyright reserved MATHEMATICS P =-1 |-3 =4 2 Which learner's response is incorrect? Hence, explain why the learner's response is incorrect. 2.3 Which term of the sequence 8; 6:: 9 will be the first to be less than ECSEYEMEIR 225 100 (2) (4) (4) Please turm ote [Page 5] QUESTION 3 3.1 3.2 The following infornmation about a quadratic number pattern is given: " T,-7= -4 " T-T. =-3 " Fourth term is cqual to l 3.1.1 3.1.2 3.1.3 3.2.1 3.2.2 Show that the general term of the quadratic number patlern is Gn+1s Determine the value of 7;, Given: S, = 2n -6n MATHEMATICS PI Which TWO terms in the quadratic number pattern will have a difference of 45? Calculate the sum of the first thirty terms. Copyright reserved Determine the value of n if 7, =300 (4) (|) (3) (2) 5 (3) |13] Please turn over
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Matched memo: 2025 G12 EC P1_V1 Sept Memo.pdf
Relevant memo page(s): 5
[Page 5] VRAAG 2 2.1 2.1.1 Amanda ✓ answer/antwoord (1) 2.1.2 For a series to converge the ratio ( )r , 1 1 r -./ Vir 'n reeks om te konvergeer moet verhouding, 1 1 r - Amanda's r is out of the interval 1 1 r -. / Amanda se r is buite die interval 1 1 r - Formula S cannot be used in such a case./ Formule S geld nie in so geval nie. ✓ r is out of the interval ( ) 1;1 r- / r is buite die interval ( ) 1;1 r- ✓ S cannot be used in such a case/ S kan dus nie gebruik word nie (2) 2.2 ( ) ( ) ( ) ( ) ( ) ( ) 1 2 3 10 3 3 3 3 2 2 2 ... 2 2 2 2 2 3 6 12 ... 3 and/ 2 1 1 3 2 1 2 1 3069 m n n a en r a r S r + + + + + + + = = - = - - = - = ✓ 3 a= ✓ 2 r = ✓ substitution/vervanging ✓ answer/antwoord (4) 2.3 ( ) 1 1 1 3 1 8 4 100 3 1 4 800 3 1 1 log log 4 800 1 log 3 800 1 ;since/ log 0 3 4 log 4 1 23,236... 24,236 25 term will thus
Free State2025SeptemberProvincial PaperEnglishMemo foundArithmeticQuadraticGeometric
2025 G12 Free State P1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 Free State P1 Sept.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 2 The general term of a quadratic number pattern is 2 n T an bn c = + + and its first term is 8. The general term of the first differences of the pattern is 4 2 kT k = - 2.1 Determine the next two terms of the number pattern nT (2) 2.2 Hence, or otherwise, show that the general term of the quadratic number pattern is given by 2 2 4 10 n T n n = - + (3) 2.3 Which term of the quadratic number pattern will be equal to 3050? (3) [8] [Page 4] QUESTION 3 3.1 Given the geometric sequence: 9 2 ; 9; 18; .. . ; 2304 3.1.1 Determine the value of r , the common ratio. (1) 3.1.2 Does the sequence converge? Motivate your answer. (2) 3.1.3 How many terms are there in the sequence? (3) 3.1.4 Determine the sum of the first 10 terms of the sequence. (2) 3.2 If the first term of the arithmetic series is 18, the common difference is 3 and the last term is 99. 3.2.1 Write down the next two terms of the series. (2) 3.2.2 Calculate the sum of the series. (5) 3.2.3 If the even numbers are removed from the series, calculate the sum of the remaining terms in the series. (5) [20]
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Matched memo: 2025 G12 Free State P1_V1 Sept Memo.pdf
Relevant memo page(s): 5
[Page 5] Mathematics P1/Wiskunde V1 5 FS/VS/September 2025 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief QUESTION/VRAAG 2 2.1 8 ; ; 10 16 ; 26 40 2 6 10 14 4 4 4 10; 16 ✓ 10; ✓16 (2) 2.2 ( ) ( ) 2 4 2 3 2 3 2 2 4 8 2 4 8 10 a a a b b b a b c c c = = + = + = = - + + = + - + = = 2a= 4 3(2) + b= 2 2 + (-4) + c= 8 (3) 2.3 ( )( ) 2 2 2 4 10 3050 2 1520 0 40 38 0 n n n n n n - + = - - = - + = n= 40 or/of n≠-38 ✓Equating (A) ✓Factors n= 40 with rejection (3) [8]
Gauteng2025SeptemberProvincial PaperEnglishMemo foundQuadraticGeometric
2025 G12 GP P1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 GP P1 Sept.pdf
Relevant question page(s): 5, 6, 7, 9, 10
[Page 5] QUESTION 2 The following sequence of numbers forms a quadratic number pattern: ;3 ;2 ;3 ;6 ; 11 ... 2.1 The first differences of the given sequence also form a sequence. Determine an expression for the th n term of the first differences. (3) 2.2 Calculate the first difference between the th 35 and th 36 terms of the quadratic sequence. (1) [Page 6] MATHEMATICS (Paper 1) 10611/25 6 P.T.O. 2.3 Determine an expression for the th n term of the quadratic sequence. (4) 2.4 Explain why the sequence will never contain a POSITIVE term. (2) [10]
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Matched memo: 2025 G12 GP P1 Sept Memo.pdf
Relevant memo page(s): 8
[Page 8] 10611/25 8 QUESTION 2 2.1 ;3 - ;2 - ;3 - ;6 - ; 11 - ... first difference: ;1 + ;1 - ;3 - ;5 - ... 1 = a and 2 - = d 3 2 2 2 1 ) 2 )( 1 ( 1 )1 ( + - = + - = - - + = - + = n T n T n T d n a T n n n n ✓ st 1 difference ✓ substitute into correct formula ✓ answer (3) 2.2 67 3 ) 35 ( 2 3 2 35 35 - = + - = + - = T T n Tn NOTE: Substitution must be n = 35. Answer only, full marks. ✓ answer (1) 2.3 6 4 6 4 2 )1( 4 )1( ) 2 ( ) 2 ( 2 2 2 4 2 ) 2 ( ) 2 ( )1( 2 1 3 )1( )1( 1 2 2 2 2 2 2 1 2 2 - + - = - = + = - = - + = + + - = - + + - = + = - + + - = - + + - = + + - = - = - = + + = n n T c c in b c b c b c b T c b c b c b T c bn n T a a but c bn an T n n n ✓ nd 2 difference ✓ value of a ✓ value of b ✓ value of c (4)
Gauteng2025SeptemberProvincial PaperAfrikaansMemo foundQuadraticGeometric
2025 G12 GP V1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 GP V1 Sept.pdf
Relevant question page(s): 5, 6, 7
[Page 5] VRAAG 2 Die volgende ry getalle vorm ʼn kwadratiese getalpatroon: ;3 ;2 ;3 ;6 ; 11 ... 2.1 Die eerste verskille van hierdie patroon vorm ook ʼn patroon. Bepaal ʼn uitdrukking vir die de n term van die eerste verskille. (3) 2.2 Bereken die eerste verskil tussen die 35ste en 36ste terme van die kwadratiese getalpatroon. (1) [Page 6] WISKUNDE (Vraestel 1) 10611/25 6 b.o. 2.3 Bepaal ʼn uitdrukking vir die de n term van die kwadratiese getalpatroon. (4) 2.4 Verduidelik hoekom die patroon nooit ʼn POSITIEWE term sal bevat nie. (2) [10]
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Matched memo: 2025 G12 GP V1 Sept Memo.pdf
Relevant memo page(s): 8
[Page 8] 10611/25 8 VRAAG 2 2.1 ;3 - ;2 - ;3 - ;6 - ; 11 - ... Eerste verskil: ;1 + ;1 - ;3 - ;5 - ... 1 = a en 2 - = d 3 2 2 2 1 ) 2 )( 1 ( 1 )1 ( + - = + - = - - + = - + = n T n T n T d n a T n n n n ✓ 1ste verskil ✓ vervang in die regte formule ✓ antwoord (3) 2.2 67 3 ) 35 ( 2 3 2 35 35 - = + - = + - = T T n Tn LET WEL: Substitusie moet weees n = 35. Volpunte vir antwoord alleenlik. ✓ antwoord (1) 2.3 2 2 2 1 2 2 2 2 2 1 (1) (1) 3 1 2 (1) (2) (2) 2 4 2 2 2 (2) (2) (1) 4 (1) 2 4 6 4 6 n n n T an bn c maar a a T n bn c T b c b c b c T b c b c b c b in c c T n n = + + = - = - = - + + = - + + -= -+ + -= + = - + + -= -+ + = + - = -= + = - = - + - ✓ 2de verskil ✓ waarde van a ✓ waarde van b ✓ waarde van c (4)
General Source2025SeptemberProvincial PaperEnglishMemo foundArithmeticSigma
2025 G12 WC P1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 WC P1 Sept.pdf
Relevant question page(s): 3, 4
[Page 3] QUESTION 2 2.1 The following arithmetic sequence is given: -7 ; -1 ; 5 ; ... . . ; 167 2.1.1 How many terms are there in this sequence? (2) 2.1.2 Calculate the value of n for which the sum of n terms will be 8208. (4) 2.2 Given the series: (x) + (x)2 2 + (x)3 4 + . ... 2.2.1 Determine the nth term in terms of x. (2) 2.2.2 Determine the value(s) of x for which the series will converge. (2) 2.2.3 If the sum of the first two terms is equal to 5 8 and x> 0, determine S∞. (6) [Page 4] 4 2.3 Given 2.3.1 Calculate the first term of the series. (1) 2.3.2 For which values of x will exist? (3) [20] QUESTION 3 Tanya starts a fitness programme by going for a run on Saturday in the local park. On the first Saturday, she runs 1 km and plans to increase her distance by 0,75 km every Saturday. She further decides that once she reaches the distance of 10km she will continue to run 10km every week thereafter. 3.1 Calculate the distance that Tania will run on her 9th run? (2) 3.2 Calculate the total distance that Tania will run over the first 24 Saturdays. (5) [7] QUESTION 4 Given: h(x) = 2 x-3 -4 4.1 Determine the equation of the asymptotes. (2) 4.2 Determine the x-intercept. (2) 4.3 Sketch the graph of h. Clearly label ALL intercepts with the axes and any asymptotes. (4) 4.4 Determine the values of x
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Matched memo: 2025 G12 WC P1_V1 Sept Memo.pdf
Relevant memo page(s): 5
[Page 5] 5 QUESTION 2 / VRAAG 2 2.1.1 Tn= a+ (n-1)d 167 = -7 + (n-1)(6) 167 = 6n-13 6n= 180 n= 30 ✓sub into correct formula ✓answer (2) 2.1.2 Sn= n 2 [2a+ (n-1)d] 8208 = n 2 [2(-7) + (n-1)(6)] 16 416 = 6n2 -20n 6n2 -20n-16 416 = 0 (3n+ 152)(2n-108) = 0 n≠- 152 3 or n= 54 ✓ sub into correct formula ✓ std form ✓ factorization ✓ both answers with selection (4) 2.2.1 Tn= arn-1 Tn= x(x 2) n-1 ✓ r ✓ answer (2) 2.2.2 -1 < r< 1 -1 < x 2 < 1 -2 < x< 2 ✓condition ✓answer (2) 2.2.3 x+ x2 2 = 5 8 8x+ 4x2 = 5 4x2 + 8x-5 = 0 (2x-1)(2x+ 5) = 0 x= 1 2 or x≠- 5 2 1 2 + 1 8 + 1 32 + ⋯. S∞= a 1 -r = 1 2 1 -1 4 = 2 3 ✓ equation ✓ std from ✓ answer with selection ✓ expansion ✓ sub ✓ answer (6) 2.3.1 (2x-1)2 or 4x2 -4x+ 1 ✓ answer (1) 2.3.2 r= 2x-1 2x-1 ≠1 x≠1 ✓ r ✓ r≠1 ✓ answer (3) [20]
General Source2025SeptemberProvincial PaperAfrikaansMemo foundArithmeticSigmaConvergence
2025 G12 WC V1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 WC V1 Sept.pdf
Relevant question page(s): 3, 4
[Page 3] VRAAG 2 2.1 Die volgende rekenkundige reeks word gegee: -7 ; -1 ; 5 ; ... . . ; 167 2.1.1 Hoeveel terme is daar in hierdie reeks? (2) 2.1.2 Bereken die waarde van n waarvoor die som van n terme 8208 sal wees. (4) 2.2 Gegee die reeks: (x) + (x)2 2 + (x)3 4 + . ... 2.2.1 Bepaal die nde term in terme van x. (2) 2.2.2 Bepaal die waarde(s) van x waarvoor die reeks sal konvergeer. (2) 2.2.3 As die som van die eerste twee terme gelyk is aan 5 8 en x> 0, bepaal die S∞. (6) [Page 4] 4 2.3 Gegee 2.3.1 Bereken die eerste term van die reeks. (1) 2.3.2 Vir watter waardes van x sal bestaan? (3) [20] VRAAG 3 Tanya begin 'n fiksheid program deur Saterdag in die plaaslike park te gaan draf. Op die eerste Saterdag hardloop sy 1 km en beplan om haar afstand elke Saterdag met 0,75 km te vermeerder. Sy besluit verder dat sodra sy die afstand van 10 km bereik, sy daarna elke week 10 km sal aanhou hardloop. 3.1 Bereken die afstand wat Tania op haar 9de draf sal aflê? (2) 3.2 Bereken die totale afstand wat Tania sal draf in haar eerste 24 Saterdae. (5) [7] VRAAG 4 Gegee: h(x) = 2 x-3 -4 4.1 Bepaal die vergelyking van die asimptote. (2) 4.2 Bepaal die x-afsnit. (2) 4.3 Skets die grafiek van h. Benoem duidelik ALLE afsnitte met die asse en enige asimptote. (4) 4.4 Bepaal die waardes van x waarvoor 2 x-3
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Matched memo: 2025 G12 WC P1_V1 Sept Memo.pdf
Relevant memo page(s): 5
[Page 5] 5 QUESTION 2 / VRAAG 2 2.1.1 Tn= a+ (n-1)d 167 = -7 + (n-1)(6) 167 = 6n-13 6n= 180 n= 30 ✓sub into correct formula ✓answer (2) 2.1.2 Sn= n 2 [2a+ (n-1)d] 8208 = n 2 [2(-7) + (n-1)(6)] 16 416 = 6n2 -20n 6n2 -20n-16 416 = 0 (3n+ 152)(2n-108) = 0 n≠- 152 3 or n= 54 ✓ sub into correct formula ✓ std form ✓ factorization ✓ both answers with selection (4) 2.2.1 Tn= arn-1 Tn= x(x 2) n-1 ✓ r ✓ answer (2) 2.2.2 -1 < r< 1 -1 < x 2 < 1 -2 < x< 2 ✓condition ✓answer (2) 2.2.3 x+ x2 2 = 5 8 8x+ 4x2 = 5 4x2 + 8x-5 = 0 (2x-1)(2x+ 5) = 0 x= 1 2 or x≠- 5 2 1 2 + 1 8 + 1 32 + ⋯. S∞= a 1 -r = 1 2 1 -1 4 = 2 3 ✓ equation ✓ std from ✓ answer with selection ✓ expansion ✓ sub ✓ answer (6) 2.3.1 (2x-1)2 or 4x2 -4x+ 1 ✓ answer (1) 2.3.2 r= 2x-1 2x-1 ≠1 x≠1 ✓ r ✓ r≠1 ✓ answer (3) [20]
KwaZulu-Natal2025SeptemberProvincial PaperEnglishMemo foundArithmeticQuadraticGeometric
2025 G12 KZN P1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 KZN P1 Sept.pdf
Relevant question page(s): 4
[Page 4] QUESTION 2 2.1 Consider the arithmetic sequence: 3 ; 2 1 ; 4 1. . . x x x 2.1.1 Determine the value of x. (3) 2.1.2 Calculate the numerical value of the th 7 term. (3) 2.2 Consider the quadratic sequence: 2 ; 7 ; 16 ; 29 ; . . . 2.2.1 Determine the 5th term of the sequence. (1) 2.2.2 Determine the nth term of the quadratic sequence. (4) 2.2.3 Show that the sum of the first differences of the quadratic sequence can be given by: 2 S 2 3 n n n (3) 2.2.4 If the sum of the first 40 first-differences in question 2.2.3 equals 3320 (that is 40 3320 S ), which term in the quadratic sequence has a value of 3322? (2) [16] QUESTION 3 3.1 Determine the number of terms in the following geometric sequence: 1 3 3 81 3 ; ; ; ... ; 2 2 2 2 (4) 3.2 Solve for p if 7 0 1 2 9 27 3 m k k m p and 1 1 p . (6) [10]
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Matched memo: 2025 G12 KZN P1 Sept Memo.pdf
Relevant memo page(s): 5
[Page 5] QUESTION 2 2.1.1 2 1 3 2 T T T T - = - ( ) ( ) 2 1 3 4 1 2 1 x x x x + - - = -- + 4 2 2 x x + = - 6 x = { A ( ) ( ) 2 1 3 4 1 2 1 x x x x + - - = -- + CA simplification CA answer (3) 2.1.2 3 ; 13 ; 23 ; ... 3 ; 10 a d = = 7 6 T a d = + ( ) 3 6 10 = + 63 = CA arithmetic sequence CA substitution CA answer (3) 2.2.1 5 46 T = A answer (1) 2.2.2 2 4 a = 2 a = 3 5 a b + = ( ) 3 2 5 b + = 1 b = - 2 a b c + + = 2 1 2 c -+ = 1 c = 2 2 1 nT n n = -+ A value of a CA value of b CA value of c CA answer (4) 2.2.3 1st differences: 5 ; 9 ; 13 ; ... 5 ; 4 a d = = ( ) 2 1 2 n n S a n d = + - ( ) ( ) 2 5 1 4 2 n n = + - 4 6 2 n n = + 2 2 3 n n = + OR 4 1 nT n = + 2 n n S a l = + 5 4 1 2 n n S n = + + 4 6 2 n n = + 2 2 3 n n = + A substitution of 5 a = A substitution of 4 d = A 4 6 2 n n + (3) OR A substitution of 5 a = A substi
KwaZulu-Natal2025SeptemberProvincial PaperAfrikaansMemo foundArithmeticQuadraticGeometric
2025 G12 KZN V1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 KZN V1 Sept.pdf
Relevant question page(s): 4
[Page 4] VRAAG 2 2.1 Beskou die rekenkundige ry: 3 ; 2 1 ; 4 1. . . x x x - + - 2.1.1 Bepaal die waarde van x. (3) 2.1.2 Bereken die numeriese waarde van die 7de term. (3) 2.2 Beskou die kwadratiese ry: 2 ; 7 ; 16 ; 29 ; . . . 2.2.1 Bepaal die 5de term van die ry. (1) 2.2.2 Bepaal die nde term van die kwadratiese ry. (4) 2.2.3 Toon aan dat die som van die eerste verskille van die kwadratiese ry gegee kan word deur: 2 S 2 3 n n n = + (3) 2.2.4 As die som van die eerste 40 eerste verskille in vraag 2.2.3 gelyk is aan 3320 (m.a.w 40 3320 S = ), watter term in die kwadratiese ry se waarde is 3322? (2) [16] VRAAG 3 3.1 Bepaal hoeveel terme in die volgende meetkundige ry is: 1 3 3 81 3 ; ; ; ... ; 2 2 2 2 (4) 3.2 Los op vir p as ( ) 7 0 1 2 9 27 3 m k k m p = = = - - en 1 1 p - . (6) [10]
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Matched memo: 2025 G12 KZN P1 Sept Memo.pdf
Relevant memo page(s): 5
[Page 5] QUESTION 2 2.1.1 2 1 3 2 T T T T - = - ( ) ( ) 2 1 3 4 1 2 1 x x x x + - - = -- + 4 2 2 x x + = - 6 x = { A ( ) ( ) 2 1 3 4 1 2 1 x x x x + - - = -- + CA simplification CA answer (3) 2.1.2 3 ; 13 ; 23 ; ... 3 ; 10 a d = = 7 6 T a d = + ( ) 3 6 10 = + 63 = CA arithmetic sequence CA substitution CA answer (3) 2.2.1 5 46 T = A answer (1) 2.2.2 2 4 a = 2 a = 3 5 a b + = ( ) 3 2 5 b + = 1 b = - 2 a b c + + = 2 1 2 c -+ = 1 c = 2 2 1 nT n n = -+ A value of a CA value of b CA value of c CA answer (4) 2.2.3 1st differences: 5 ; 9 ; 13 ; ... 5 ; 4 a d = = ( ) 2 1 2 n n S a n d = + - ( ) ( ) 2 5 1 4 2 n n = + - 4 6 2 n n = + 2 2 3 n n = + OR 4 1 nT n = + 2 n n S a l = + 5 4 1 2 n n S n = + + 4 6 2 n n = + 2 2 3 n n = + A substitution of 5 a = A substitution of 4 d = A 4 6 2 n n + (3) OR A substitution of 5 a = A substi
Limpopo2025SeptemberProvincial PaperEnglishMemo foundPatterns & Sequences
2025 G12 Limpopo P1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 Limpopo P1 Sept.pdf
Relevant question page(s): Not isolated from scan
Topic section could not be extracted cleanly from this PDF scan, but this source paper is still included from the local Grade 12 resource bank.
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Matched memo: 2025 G12 Limpopo P1 Pre-Trial Memo.pdf
Relevant memo page(s): 5
[Page 5] 50 nT n n n equating/gelykstel substitution/vervanging answer/antwoord (3) 2.2 2.2.1 8 4 24 x x 8 4 20 x x x 4 2 2 16 x x 2 4 2 16 2 2 16 4 4 20 5 x x x x x x second difference/2e verskil equating/gelykstel 5 x (3)
Mpumalanga2025SeptemberProvincial PaperAfrikaansMemo foundPatterns & Sequences
2025 G12 Mpumalanga V1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 Mpumalanga V1 Sept.pdf
Relevant question page(s): Not isolated from scan
Topic section could not be extracted cleanly from this PDF scan, but this source paper is still included from the local Grade 12 resource bank.
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Matched memo: 2025 G12 Mpumalanga P1_V1 Sept Memo.pdf
Relevant memo page(s): Not isolated
A matching memo file was found, but the answer section could not be isolated cleanly from the current PDF scan.
North West2025SeptemberProvincial PaperEnglishMemo foundPatterns & Sequences
2025 G12 NW P1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 NW P1 Sept.pdf
Relevant question page(s): Not isolated from scan
Topic section could not be extracted cleanly from this PDF scan, but this source paper is still included from the local Grade 12 resource bank.
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Matched memo: 2025 G12 NW P1 Memo.pdf
Relevant memo page(s): Not isolated
A matching memo file was found, but the answer section could not be isolated cleanly from the current PDF scan.
North West2025SeptemberProvincial PaperAfrikaansMemo not linkedQuadraticGeometric
2025 G12 NW V1 Sept
Source folder: 2025 G12 Provincial June Exams
Source file: 2025 G12 NW V1 Sept.pdf
Relevant question page(s): 3, 4
[Page 3] VRAAG 2 Gegee is die kwadratiese getalpatroon: 2; 7; 16; . . . . . . ; 862 2.1 Skryf die 4de term van die kwadratiese getalpatroon neer. (1) 2.2 Bepaal die algemene term ) (Tn vir die kwadratiese getalpatroon. (4) 2.3 Bereken die aantal terme in die getalpatroon. (3) [8] [Page 4] VRAAG 3 3.1 Gegee die meetkundige reeks: 2 + 6 + 18 + 54 + . . . 3.1.1 Skryf die algemene term van hierdie reeks neer. (1) 3.1.2 Bereken die waarde van m sodanig dat: m n n 1 048 59 3 3 2 (4) 3.2 Die eerste term van 'n oneindige meetkundige ry is 3 en die gemeenskaplike verhouding van die ry is 1 2. 3.2.1 Bepaal die waarde van die derde term van die ry. (1) 3.2.2 Bepaal die waarde van S∞. (2) 3.3 Indien die magte van 6 uit die ry verwyder word 1; 2; 3; 4; . . . ; 8 000, bepaal die som van die oorblywende terme. (4) [12]
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Northern Cape2025SeptemberProvincial PaperEnglishMemo foundArithmeticGeometricSigma
2025 G12 NC P1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 NC P1 Sept.pdf
Relevant question page(s): 4, 5, 10
[Page 4] QUESTION 2 2.1 Consider the arithmetic sequence: 15 ; 13 ; 11 ; 9 ; ... ; -121 2.1.1 Calculate the number of terms in the sequence. (3) 2.1.2 Hence, calculate the value of the following series: 7 + 15 + 7 + 13 + 7 + 11 + 7 + 9 + 7 ... + 7 - 121 (4) 2.2 Consider the following sequence of numbers illustrated by the columns and rows below. 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 Row 1 Row 2 Row 3 Row 4 Row 5 Row 6 2.2.1 The first term of Row n is given by Tn= an2 + bn+ c. Determine the values of a, b and c. (3) 2.2.2 Hence, calculate the first term of Row 21. (2) 2.2.3 Calculate the 10th term of Row 21. (2) [14] [Page 5] QUESTION 3 3.1 Consider the geometric series: 729 + 243 + 81 + ... 3.1.1 Calculate T9. (3) 3.1.2 It is given that: 729 + 243 + 81 + ... + Tm = 1 10933 Calculate m. (3) 3.2 Consider the following infinite geometric series: 1 + r+ r2 + r3 +... The second and each consecutive term that follows, is twice the sum of all the terms in the series following that term. Determine the value of the 2nd and 3rd terms of the series. (5) [11]
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Matched memo: 2025 G12 NC P1_V1 Sept Memo.pdf
Relevant memo page(s): 4
[Page 4] an 17 and/en d into correct formula/in korreke formule equating to -121/gelykstelling aan -121 n (3) 2.1.2 69 2 ( 1) 2 69 2(15) 68( 2) 2 3657 3657 69(7) 3174 n series n S a n d S S = + - = + - = - = - + = - OR/OF 69 2 69 15 ( 121) 2 3657 3657 69(7) 3174 n series n S a l S S = + = + - = - = - + = - subst into correct formula/subst. in korrekte formule S69 69(7) answer/antwoord OR/OF subst into correct formula/subst. in korrekte formule S69 69(7) answer/antwoord (4)
Northern Cape2025SeptemberProvincial PaperAfrikaansMemo foundArithmeticGeometricSigma
2025 G12 NC V1 Sept
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 NC V1 Sept.pdf
Relevant question page(s): 4, 5, 10
[Page 4] VRAAG 2 2.1 Beskou die rekenkundige ry: 15 ; 13 ; 11 ; 9 ; ... ; -121 2.1.1 Bereken die aantal terme in die ry. (3) 2.1.2 Bereken vervolgens die waarde van die volgende reeks: 7 + 15 + 7 + 13 + 7 + 11 + 7 + 9 + 7 ... + 7 - 121 (4) 2.2 Beskou die rye getalle wat deur die kolomme en rye hieronder voorgestel word. 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 Ry 1 Ry 2 Ry 3 Ry 4 Ry 5 Ry 6 2.2.1 Die eerste term van Ry n word gegee deur Tn= an2 + bn+ c. Bepaal die waardes van a, b en c. (3) 2.2.2 Bereken vervolgens die eerste term van Ry 21. (2) 2.2.3 Bereken die 10de term van Ry 21. (2) [14] [Page 5] VRAAG 3 3.1 Beskou die meetkundige reeks: 729 + 243 + 81 + ... 3.1.1 Bereken T9. (3) 3.1.2 Dit word gegee dat: 729 + 243 + 81 + ... + Tm = 1 10933 Bereken m. (3) 3.2 Beskou die volgende oneindige meetkunde reeks: 1 + r+ r2 + r3 +... Die tweede en elke opeenvolgende term wat volg, is twee keer die som van al die terme in die reeks wat die term volg. Bepaal die waarde van die 2de en 3de terme van die reeks. (5) [11]
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Matched memo: 2025 G12 NC P1_V1 Sept Memo.pdf
Relevant memo page(s): 4
[Page 4] an 17 and/en d into correct formula/in korreke formule equating to -121/gelykstelling aan -121 n (3) 2.1.2 69 2 ( 1) 2 69 2(15) 68( 2) 2 3657 3657 69(7) 3174 n series n S a n d S S = + - = + - = - = - + = - OR/OF 69 2 69 15 ( 121) 2 3657 3657 69(7) 3174 n series n S a l S S = + = + - = - = - + = - subst into correct formula/subst. in korrekte formule S69 69(7) answer/antwoord OR/OF subst into correct formula/subst. in korrekte formule S69 69(7) answer/antwoord (4)
SACAI2025NovemberSACAI PaperEnglishMemo foundArithmeticQuadraticGeometricSigma
2025 G12 SACAI P1 Nov
Source folder: 2024 - 2025 SACAI Exams
Source file: 2025 G12 SACAI P1 Nov.pdf
Relevant question page(s): 4, 16
[Page 4] QUESTION 2 2.1 The following arithmetic sequence is given: 2; 9; 16; .... Calculate the 100th term in this sequence. (3) 2.2 Evaluate the following arithmetic series: 1 -1 -3 -⋯-193. (4) 2.3 The sum of an infinite, converging geometric series, is 10. The sum to infinity of the odd numbered terms of the same series is 25 3 . Determine the first term of this series. (6) [13] QUESTION 3 3.1 Solve for x∈N if: ∑1 3 (2k-1) x k=1 > 84 (4) 3.2 A train traveling between two cities stops at various stations. The table below shows the number of passengers on the train at each stop: STOP 1 STOP 2 STOP 3 STOP 4 Number of passengers 7 28 43 52 The number of passengers on the train forms a quadratic sequence, Tn= an2 + bn+ c. 3.2.1 Determine the values of a, b and c. (3) 3.2.2 If Tn= -3n2 + 30n-20, calculate the maximum num [Page 16] VRAAG 2 2.1 Die volgende rekenkundige ry is gegee: 2; 9; 16; .... Bereken die 100ste term in hierdie ry. (3) 2.2 Evalueer die volgende rekenkundige reeks: 1 -1 -3 -⋯-193. (4) 2.3 Die som van ʼn oneindige konvergerende meetkundige reeks is 10. Die som tot oneindig van die onewe genommerde terme van dieselfde reeks is 25 3 . Bepaal die eerste term van hierdie reeks. (6) [13] VRAAG 3 3.1 Los op vir x∈N as: ∑1 3 (2k-1) x k=1 > 84 (4) 3.2 A trein wat tussen twee stede ry, stop by verskeie stasies. Die tabel hieronder toon die aantal passasiers op die trein na elke stop: STOP 1 STOP 2 STOP 3 STOP 4 Aantal passasiers 7 28 43 52 Die aantal passasiers op die trein vorm ʼn kwadratiese ry, Tn= an2 + bn+ c. 3.2.1 Bepaal die waardes van a, b en c. (3) 3.2.2 As Tn= -3n2 + 30n-20, bereken die maksimum aantal passasiers op d
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Matched memo: 2025 G12 SACAI P1 Nov Memo.pdf
Relevant memo page(s): 5
[Page 5] Subst. into correct formula ✔ -192 ✔ Answer (4) 2.3 S∞= 10 = a 1-r (1) 25 3 = a 1-r2 (2) (1) ÷ (2): 6 5 = (1-r)(1+r) 1-r 6 5 = 1 + r r= 1 5 10 = a 1-1 5 a= 8 ✔ Sum to infinity ✔ r2 ✔ Calculation ✔ Simplify ✔ Answer r= 1 5 ✔ Answer a (6) [13]
General Source2025AssignmentAssignmentMemo not linkedArithmeticQuadraticGeometricConvergence
2025 G12 Term 2 Opdrag 2025
Source folder: Assignments
Source file: 2025 G12 Term 2 Opdrag 2025.pdf
Relevant question page(s): 2, 3, 4
[Page 2] VRAAG 2 2.1 Gegee die meetkundige reeks: 2.1.1 Vir watter waarde(s) van sal die reeks konvergeer? (3) 2.1.2 Bepaal die as (2) 2.2 Die tweede term van 'n kwadratiese getalpatroon is 13. Die algemene term van die eerste verskil van die kwadratiese getalpatroon word gegee deur . 2.2.1 Bepaal die eerste 5 terme van die kwadratiese getalpatroon. (3) 2.2.2 Bepaal dus die algemene term (4) [Page 3] 2.3 Gegee: 2; 3; 5; 6; 8; 12; ... 2.3.1 Bepaal die volgende twee terme in die ry. (2) 2.3.2 Bepaal die som van die eerste 39 terme. (5) 2.4 Skryf die volgende reeks in notasie. tot terme. (3) 2.5 Die getalle vorm 'n meetkundige ry. As die som van hierdie drie getalle is, bepaal die moontlike waarde(s) van en . (6) 2.6 Die volgende meetkundige reeks is gegee: 2.6.1 Bepaal die waarde van die konstante verhouding, in terme van . (1) 2.6.2 As , bereken die som van die reeks vir die eerste 10 terme. (3) 2.7 Die som van die eerste terme van 'n rekenkundige reeks word gegee deur , 2.7.1 Bereken die som van die eerste 15 terme. (2) 2.7.2 Bereken die waarde van (2) 2.7.3 As die eerste term van die ry 7 is, watter term van die ry sal 'n waarde van hê? (4) 2.8 Gegee dat: 2.8.1 Bereken die waardes van waarvoor sal konve
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DBE / National2025ExemplarExemplarAfrikaansMemo not linkedArithmeticSigma
2025 G12 DBE V1 Exemplar
Source folder: 2025 G12 Provincial Sept Exams
Source file: 2025 G12 DBE V1 Exemplar.pdf
Relevant question page(s): 3
[Page 3] VRAAG 2 2.1 Die eerste term van 'n rekenkundige reeks is 7. Die gemeenskaplike verskil van hierdie reeks is 5 en die reeks bevat 20 terme. 2.1.1 Bereken die som van hierdie reeks. (2) 2.1.2 Die oorspronklike rekenkundige reeks word na 75 terme uitgebrei. Die som van hierdie 75 terme is 14 400. Gebruik sigma-notasie om 'n vergelyking vir die som van die terme wat by die oorspronklike reeks getel is, neer te skryf. (4) 2.2 Die ry van die eerste verskille van 'n kwadratiese patroon is: 1 ; 3 ; 5 ; ... 2.2.1 Indien 99 T van die kwadratiese patroon 9 632 is, bereken die waarde van 98 T . (3) 2.2.2 Indien daar verder gegee word dat die derde term van die kwadratiese patroon 32 is, bepaal die algemene term, n T , van die kwadratiese patroon. (5) [14]
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General Source2025ExemplarExemplarEnglishMemo not linkedArithmeticQuadraticSigma
Mathematics_P1_Gr_12_Exemplar_2025_Eng
Source folder: Maths Paper 1 Exemplar Paper with Answer Book
Source file: Mathematics_P1_Gr_12_Exemplar_2025_Eng.pdf
Relevant question page(s): 3
[Page 3] QUESTION 2 2.1 The first term of an arithmetic series is 7. The common difference of this series is 5 and the series contains 20 terms. 2.1.1 Calculate the sum of this series. (2) 2.1.2 The original arithmetic series is extended to 75 terms. The sum of these 75 terms is 14 400. Using sigma notation, write down an equation for the sum of the terms added to the original series. (4) 2.2 The sequence of the first differences of a quadratic pattern is: 1 ; 3 ; 5 ; ... 2.2.1 If 99 T of this quadratic pattern is 9 632, calculate the value of 98 T . (3) 2.2.2 If it is further given that the third term of the quadratic pattern is 32, determine the general term, n T , of the quadratic pattern. (5) [14]
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