S6 Mathematics Paper: Arithmetic And Geometric Sequences And Summation (100 Marks)

S6 Mathematics Paper: Arithmetic and Geometric Sequences and Summation (100 marks)

2026-11-12

Mathematics – Sequences and Summation

S6 Maths: 6 questions, 24 parts, 100 marks, 75 min. Arithmetic and geometric sequences, summation, compound interest, depreciation, means and terms from partial sums.

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Section A: Arithmetic Sequences – General Term and Partial Sums (17 marks)

An arithmetic sequence has first term 7 and common difference 5. The nth term is T(n) = a + (n – 1)d and the sum of the first n terms is S(n) = n/2 [2a + (n – 1)d].

#	Question	Answer and marking notes
1	(a) Find the 20th term of the sequence. (4 marks)	Answer: T(20) = 7 + (20 – 1) x 5 = 7 + 95 = 102. Marking: 2 marks for substituting into the general term, 1 for the arithmetic, 1 for the answer.
2	(b) Find the sum of the first 15 terms. (4 marks)	Answer: S(15) = 15/2 [2 x 7 + 14 x 5] = 15/2 x 84 = 630. Marking: 2 marks for the formula, 1 for the bracket, 1 for the answer.
3	(c) If one term of the sequence equals 202, find its position. (4 marks)	Answer: 7 + (n – 1) x 5 = 202, so (n – 1) x 5 = 195, n – 1 = 39 and n = 40; 202 is the 40th term. Marking: 2 marks for the equation, 1 for solving, 1 for the answer.
4	(d) Find the smallest n for which the sum of the first n terms exceeds 2000. (5 marks)	Answer: S(n) = n/2 [14 + 5(n – 1)] = n(5n + 9)/2, so we need n(5n + 9) > 4000. For n = 27, 27 x 144 = 3888, too small; for n = 28, 28 x 149 = 4172, large enough. Hence n = 28 with S(28) = 2086. Marking: 2 marks for simplifying S(n), 1 for the inequality, 1 for testing values, 1 for the answer.

Section B: Geometric Sequences – General Term, Partial and Infinite Sums (17 marks)

A geometric sequence has first term 3 and common ratio 2. T(n) = a r^(n-1), S(n) = a(r^n – 1)/(r – 1), and for |r| less than 1 the sum to infinity is a/(1 – r).

#	Question	Answer and marking notes
5	(a) Find the 10th term. (4 marks)	Answer: T(10) = 3 x 2^9 = 3 x 512 = 1536. Marking: 2 marks for the general term, 1 for the index, 1 for the answer.
6	(b) Find the sum of the first 8 terms. (4 marks)	Answer: S(8) = 3(2^8 – 1)/(2 – 1) = 3 x 255 = 765. Marking: 2 marks for the formula, 1 for the index, 1 for the answer.
7	(c) A second geometric sequence has first term 48 and common ratio 3/4. Find its sum to infinity. (4 marks)	Answer: since \|3/4\| less than 1, S(inf) = 48 / (1 – 3/4) = 48 / 0.25 = 192. Marking: 1 mark for convergence, 2 for the formula, 1 for the answer.
8	(d) For the sequence with first term 3 and ratio 2, find the smallest n with T(n) greater than 10000. (5 marks)	Answer: 3 x 2^(n-1) > 10000 gives 2^(n-1) > 3333.3. Since 2^11 = 2048 is too small and 2^12 = 4096 is enough, n – 1 = 12 and n = 13, with T(13) = 12288. Marking: 2 marks for the inequality, 2 for the comparison, 1 for the answer.

Section C: Sigma Notation and Applications of Summation (17 marks)

Useful results: sum of k from 1 to n is n(n + 1)/2; sum of a r^(k-1) is a(r^n – 1)/(r – 1); the sum of the first n odd numbers is n squared.

#	Question	Answer and marking notes
9	(a) Evaluate the sum from k = 1 to 20 of (3k + 2). (4 marks)	Answer: the sum equals 3 x (20 x 21/2) + 2 x 20 = 630 + 40 = 670. Marking: 1 mark for splitting, 1 for the sum of k, 1 for the constant term, 1 for the answer.
10	(b) Evaluate the sum from k = 1 to 12 of 2 raised to k. (4 marks)	Answer: this is a geometric sum with first term 2, ratio 2 and 12 terms: 2(2^12 – 1)/(2 – 1) = 2 x 4095 = 8190. Marking: 2 marks for recognising the geometric sum, 1 for the formula, 1 for the answer.
11	(c) An arithmetic sequence has first term 5, common difference 6 and last term 89. Find the number of terms and their sum. (4 marks)	Answer: 5 + (n – 1) x 6 = 89 gives n – 1 = 14 and n = 15; the sum is 15/2 (5 + 89) = 7.5 x 94 = 705. Marking: 2 marks for n, 1 for the sum, 1 for the answer.
12	(d) Using the fact that the first n odd numbers sum to n squared, evaluate 1 + 3 + 5 + … + 49, and explain why this agrees with the arithmetic sum formula. (5 marks)	Answer: 2n – 1 = 49 gives n = 25, so the sum is 25 squared = 625. Checking with the arithmetic formula: first term 1, difference 2, 25 terms, S = 25/2 [2 + 24 x 2] = 25/2 x 50 = 625, which agrees. Marking: 2 marks for n, 1 for the square, 2 for the check.

Section D: Compound Interest, Depreciation and Growth Models (17 marks)

Compound interest: amount = P(1 + r)^n. Reducing-balance depreciation: value = P(1 – r)^n. For equal annual deposits A at rate r over n periods the accumulated amount is A[(1 + r)^n – 1]/r.

#	Question	Answer and marking notes
13	(a) 50000 dollars is deposited at 4 per cent per year compounded yearly. Find the amount after 6 years, correct to the nearest cent. (4 marks)	Answer: 50000 x 1.04^6 = 50000 x 1.265319 = 63,265.95 dollars. Marking: 2 marks for the formula, 1 for the power, 1 for the answer with unit.
14	(b) A machine costs 200000 dollars and depreciates at 15 per cent of its book value each year. Find its book value after 4 years. (4 marks)	Answer: 200000 x 0.85^4 = 200000 x 0.52200625 = 104,401.25 dollars. Marking: 2 marks for the depreciation factor, 1 for the power, 1 for the answer.
15	(c) From (a), find the least number of years needed for the amount to reach at least 100000 dollars. (4 marks)	Answer: we need 1.04^n at least 2. Since 1.04^17 is about 1.948 and 1.04^18 is about 2.026, at least 18 years are needed. Marking: 2 marks for the inequality, 1 for testing, 1 for the answer.
16	(d) 12000 dollars is deposited at the end of each year for 10 years at 4 per cent per year. Find the accumulated amount. (5 marks)	Answer: the amount is 12000 x [(1.04^10 – 1)/0.04] = 12000 x 12.006107 = 144,073.29 dollars. Marking: 2 marks for recognising the geometric sum, 2 for the formula, 1 for the answer.

Section E: Arithmetic and Geometric Means, Sum to Infinity (16 marks)

If a, b, c are in arithmetic progression then b = (a + c)/2; if they are in geometric progression with the same sign then b squared equals ac. Inserting k arithmetic means between two numbers uses common difference (last – first)/(k + 1).

#	Question	Answer and marking notes
17	(a) An arithmetic sequence has 3rd term 14 and 7th term 30. Find its first term and common difference. (4 marks)	Answer: subtracting gives 4d = 16 so d = 4; then a + 2d = 14 gives a = 6. Marking: 2 marks for the two equations, 1 for d, 1 for a.
18	(b) If 2, x and 18 are in geometric progression and x is positive, find x. (4 marks)	Answer: x squared = 2 x 18 = 36, and x is positive so x = 6. Marking: 3 marks for the geometric mean relation, 1 for taking the positive root.
19	(c) A geometric sequence has first term 64 and common ratio -1/2. Find its sum to infinity. (4 marks)	Answer: since the modulus of -1/2 is below 1, S(inf) = 64 / (1 + 1/2) = 64 x 2/3 which is about 42.67 (that is 128/3). Marking: 1 mark for convergence, 2 for the formula, 1 for the answer.
20	(d) Insert 4 arithmetic means between 5 and 30 and state them. (4 marks)	Answer: there are 5 intervals, so the common difference is (30 – 5)/5 = 5 and the four means are 10, 15, 20, 25. Marking: 2 marks for the difference, 2 for listing the terms.

Section F: Recovering Terms from Partial Sums and Mixed Applications (16 marks)

If S(n) is the sum of the first n terms then T(n) = S(n) – S(n – 1) for n at least 2, and T(1) = S(1). A recurring decimal is the sum of an infinite geometric sequence.

#	Question	Answer and marking notes
21	(a) A sequence has S(n) = 3n squared + 5n. Find T(n), the first term and the common difference, and state whether it is arithmetic. (4 marks)	Answer: T(n) = S(n) – S(n – 1) = 3n squared + 5n minus [3(n – 1) squared + 5(n – 1)] = 6n + 2, so T(1) = 8, T(2) = 14 and the common difference is 6; T(n) is linear in n, so the sequence is arithmetic. Marking: 2 marks for the subtraction, 1 for the first term and difference, 1 for the conclusion.
22	(b) A geometric sequence has the sum of its first 3 terms equal to 26 and its sum to infinity equal to 27. Find a and r. (4 marks)	Answer: a/(1 – r) = 27 and a(1 + r + r squared) = 26, so dividing gives (1 – r)/(1 + r + r squared) = 26/27 and hence r = 1/3; then a = 27(1 – r) = 18. Marking: 1 mark for the equations, 2 for r, 1 for a.
23	(c) Express the recurring decimal 0.272727… (27 recurring) as a fraction in lowest terms. (4 marks)	Answer: 0.272727… = 0.27 + 0.0027 + … with first term 0.27 and ratio 0.01, so the sum is 0.27/(1 – 0.01) = 0.27/0.99 = 27/99 = 3/11. Marking: 1 mark for the geometric view, 2 for the sum, 1 for cancelling.
24	(d) Three numbers in geometric progression have sum 21 and product 216. Find the numbers. (4 marks)	Answer: let the terms be a/r, a and ar; the product gives a cubed = 216 so a = 6, and 6/r + 6 + 6r = 21 gives 2r squared – 5r + 2 = 0, so r = 2 or 1/2 and the numbers are 3, 6, 12. Marking: 1 mark for the setup, 1 for a, 1 for r, 1 for the answer.

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