This set has twelve original questions, ordered from easier to harder, covering all five lessons in arithmetic and geometric series. Questions 1 to 4 are arithmetic terms and sums, 5 to 7 use geometric ratios and finite sums, 8 to 10 involve a condition or a sum to infinity, and 11 and 12 are word problems.
Attempt each question on paper and check your answer against the original conditions before opening the solution. The sequence and series laboratory is useful for checking after you have tried by hand, and the mistake log and retest queue helps you track errors.
Mark the ones you got wrong and use the routing list at the end.
Questions
1. An arithmetic progression has first term 5 and common difference 3. Find the 30th term.
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30th term = 5 + 29 × 3 = 5 + 87 = 92.
Check: the 1st term is 5 and there are 29 steps of 3 after it, so 5 + 87 = 92.
2. Find the sum of the first 25 terms of 7 + 12 + 17 + …
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a = 7 and d = 5. S25 = 25/2 × (14 + 24 × 5) = 25/2 × 134 = 25 × 67 = 1675.
Check: last term = 7 + 120 = 127, and 25/2 × (7 + 127) = 25/2 × 134 = 1675.
3. The 4th term of a geometric progression is 24 and the 7th term is 192. Find r and the first term.
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ar³ = 24 and ar⁶ = 192. Dividing, r³ = 8, so r = 2. Then 8a = 24, so a = 3.
Check: 3, 6, 12, 24, 48, 96, 192, so the 4th term is 24 and the 7th term is 192.
4. The arithmetic progression 2, 9, 16, … ends with the term 93. How many terms are there, and what is their sum?
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d = 7. Solve 2 + 7(n − 1) = 93, so 7(n − 1) = 91 and n − 1 = 13. Hence n = 14.
S = 14/2 × (2 + 93) = 7 × 95 = 665. Check: the average of the first and last is 47.5, and 14 × 47.5 = 665.
5. Find the sum of the first 8 terms of the geometric progression 3, 6, 12, …
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a = 3 and r = 2. S8 = 3(2⁸ − 1)/(2 − 1) = 3 × 255 = 765.
Check by adding: 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765.
6. The 2nd term of a geometric progression is 12 and the 4th term is 48. Write down every possible pair of values of a and r.
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ar = 12 and ar³ = 48. Dividing, r² = 4, so r = 2 or r = −2.
If r = 2, then a = 6. If r = −2, then a = −6. Both work: 6, 12, 24, 48 and −6, 12, −24, 48. So (a, r) = (6, 2) or (−6, −2).
7. An arithmetic progression has 4th term 11 and the sum of its first 10 terms is 155. Find a and d.
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a + 3d = 11 and 10/2 × (2a + 9d) = 155, so 2a + 9d = 31. From the first, a = 11 − 3d. Then 22 − 6d + 9d = 31, so d = 3 and a = 2.
Check: the terms 2, 5, 8, 11 have 4th term 11. S10 = 5 × (4 + 27) = 155.
8. The numbers x, x + 2 and x + 5 are consecutive terms of a geometric progression. Find x and r.
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(x + 2)² = x(x + 5), so x² + 4x + 4 = x² + 5x and x = 4.
The terms are 4, 6, 9, so r = 3/2. Check: 6 ÷ 4 = 1.5 and 9 ÷ 6 = 1.5.
9. Find the sum to infinity of 50 − 10 + 2 − …, and say whether 5 + 7.5 + 11.25 + … has one.
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For the first series, r = −10 ÷ 50 = −1/5 and |r| < 1. S∞ = 50 ÷ (1 + 1/5) = 50 ÷ 6/5 = 250/6 = 125/3 (about 41.7).
For the second series, r = 7.5 ÷ 5 = 1.5, and |r| > 1, so it has no sum to infinity.
10. A geometric series has 2nd term 6 and sum to infinity 25. Find all possible values of r and the first term.
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ar = 6 and a/(1 − r) = 25. From the second, a = 25(1 − r). Substitute: 25r(1 − r) = 6, so 25r² − 25r + 6 = 0.
Then (5r − 2)(5r − 3) = 0, so r = 2/5 or r = 3/5. Both satisfy |r| < 1.
If r = 2/5, then a = 25 × 3/5 = 15. If r = 3/5, then a = 25 × 2/5 = 10. Check: 15 × 2/5 = 6 and 15 ÷ (3/5) = 25; 10 × 3/5 = 6 and 10 ÷ (2/5) = 25. So (a, r) = (15, 2/5) or (10, 3/5).
11. Aina saves RM100 in month 1 and RM15 more each month than the month before. How much has she saved in total after 12 months, and in which month does her deposit first exceed RM400?
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This is an arithmetic progression with a = 100 and d = 15. S12 = 12/2 × (200 + 11 × 15) = 6 × 365 = RM2190.
For the deposit, 100 + 15(n − 1) > 400 gives n − 1 > 20, so n > 21. Month 21 gives exactly RM400, which does not exceed it. Month 22 gives RM415, so the 22nd month.
12. A ball is dropped from 10 m. After each bounce it rises to 60% of the height of the previous fall. Find the total distance it travels before it comes to rest.
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The first fall is 10 m. After that, each rise and each fall covers the same height, so the distance is 10 + 2 × (6 + 3.6 + 2.16 + …).
The bracket is geometric with a = 6 and r = 0.6, so its sum is 6 ÷ 0.4 = 15. Total = 10 + 2 × 15 = 40 m.
Check: 6 ÷ 0.4 = 15, and the first few terms 6 + 3.6 + 2.16 = 11.76 are already moving towards 15.
If you got these wrong
Use the question number to find the lesson to revisit.
- Wrong term or wrong first term (1, 3, 7): read finding a term from two conditions, especially the n versus n − 1 slip.
- Sum off by a term or two (2, 4, 11): read calculating a finite arithmetic sum and recount n.
- Missing a second ratio, or a root error (3, 6, 8): read recovering a common ratio.
- Formula gives an odd answer (9, 10, 12): read using an infinite sum only when its condition holds.
- Stuck at the start of a long question (7, 10, 11): read solving a combined term-and-sum problem and write a plan before any algebra.
The non-calculator working trainer may also help if the algebra was right but the arithmetic slipped. Go back to the module overview for the full route.
If the same two or three questions stay difficult after revisiting the lesson, bring your written attempts to online one-to-one Additional Mathematics tuition, where a teacher can trace the exact step that goes wrong.