Free Series questions for AS Level Mathematics (9709), part of Pure Mathematics 1. Each question has the answer and the marking points an examiner would look for.
Generate a full Series worksheet →All Pure Mathematics 1 questionsThe expression (3x2 − 1x)6 is expanded.
Find the term independent of x.
Find the coefficient of x3.
Hence find the term independent of x in the expansion of (1 + 2x3)(3x2 − 1x)6.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 135 general term 6Cr(3x2)6−r(−1/x)r M1 r = 4 A1 135 A1 | 3 |
| (b) | −540 r = 3 M1 −540 A1 | 2 |
| (c) | −945 term from 1 × (a) plus 2 × (b) M1 −945 A1 | 2 |
In the expansion of (2 + ax)5, where a is a positive constant, the coefficient of x2 is 1 times the coefficient of x.
Find the value of a.
Hence find the coefficient of x2 in the expansion of (1 − 3x)(2 + ax)5.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1 5 × 24 a and 10 × 23 a2 B1 80a2 = 1 × 80a M1 1 A1 | 3 |
| (b) | −160 80 seen B1 80 − 3 × 80 M1 −160 A1 | 3 |
Find the coefficient of x3 in the expansion of (2 + 2x)7.
A geometric progression has first term 44 and sum to infinity 176. Find the common ratio and the sum of the first 6 terms.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 4480 7C3 × 24 × (2)3 M1 4480 A1 | 2 |
| (b) | r = 0.75, S6 = 145 a ÷ (1 − r) = S∞ M1 r = 0.75 A1 145 A1 | 3 |
The first, second and third terms of a geometric progression are respectively the first, 4th and 10th terms of an arithmetic progression with first term a and common difference d, where d ≠ 0.
Show that a = 3d.
Find the common ratio of the geometric progression.
Given that d = 4, find the sum of the first 10 terms of the arithmetic progression and the sum of the first 6 terms of the geometric progression.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | a = 3d (a + 3d)2 = a(a + 9d) M1 9d2 = 3ad A1 a = 3d since d ≠ 0 A1 | 3 |
| (b) | 2 r = (a + 3d)/a = 2 B1 | 1 |
| (c) | 300; 756 a = 12 B1 AP sum formula used M1 300 A1 GP sum = 756 A1 | 4 |
Find the first three terms in the expansion of (1 + ax)7 in ascending powers of x.
The coefficient of x2 in the expansion of (1 + ax)7(2 − x) is 399. Find the possible values of a.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1 + 7ax + 21a2x2 7ax B1 21a2x2 B1 | 2 |
| (b) | a = −3, a = 196 2 × 21a2 − 7a M1 42a2 − 7a − 399 = 0 A1 solve the quadratic M1 a = −3, 196 A1 | 4 |
A geometric progression has a sum to infinity of 24 and a second term of 4.5. Show that 24r2 − 24r + 4.5 = 0 and find the two possible values of the common ratio r.
For the larger value of r, find the sum of the first 6 terms.
A second progression has first term 1 and common ratio (2x − 1). Find the set of values of x for which it has a sum to infinity.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | r = 14 or 34 ar = 4.5 and a/(1 − r) = 24 B1 eliminate a M1 given quadratic obtained A1 solve M1 r = 14, 34 A1 | 5 |
| (b) | 19.7 a = 6 M1 19.7 A1 | 2 |
| (c) | 0 < x < 1 −1 < 2x − 1 < 1 M1 0 < x < 1 A1 | 2 |
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in