IGCSE · Practice questions · Geometric progressions and the sum to infinity

Divide, take logs, let it settle.

Six original Cambridge-style questions on geometric progressions: the nth term, the sum of n terms, finding a and r, the least number of terms using lg, when a sum to infinity exists and what it is, and a problem linking an AP and a GP.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Find $a$ and $r$, then choose the formula.

Ratio: $r = \dfrac{u_{2}}{u_{1}}$; with two terms given, write each as $ar^{\text{power}}$ and divide.
Term: $u_{n} = ar^{n-1}$. Sum: $S_{n} = \dfrac{a(1 - r^{n})}{1 - r}$.
Least $n$: isolate the power, take lg, divide by $\lg r$ (reverse the sign if $r < 1$), round UP and check.
Sum to infinity: only if $|r| < 1$; then $S_{\infty} = \dfrac{a}{1 - r}$. Say why in words.
Context: a fall of 12% a year means $r = 0.88$; a ball's rebounds are travelled twice.

Recall

[2 marks]

A geometric progression begins $64, -32, 16, \dots$ Find the common ratio and the 9th term.

$r = \dfrac{-32}{64} = -\dfrac{1}{2}$. ✓
$u_{9} = 64 \times \left(-\dfrac{1}{2}\right)^{8} = \dfrac{64}{256} = \dfrac{1}{4}$. ✓

Calculation

[4 marks]

The 3rd term of a geometric progression is 20 and the 6th term is 10.24.

(a) Find the common ratio and the first term.

(b) Explain why the progression has a sum to infinity, and find it.

(a) $ar^{2} = 20$, $ar^{5} = 10.24$. Dividing, $r^{3} = 0.512$, so $r = 0.8$. ✓
$a = \dfrac{20}{0.8^{2}} = \dfrac{20}{0.64} = 31.25$. ✓
(b) $|r| = 0.8 < 1$, so $r^{n} \to 0$ and the sums approach a limit. ✓
$S_{\infty} = \dfrac{31.25}{1 - 0.8} = \dfrac{31.25}{0.2} = 156.25$. ✓

Calculation

[4 marks]

A machine is bought for $40 000. At the end of each year its value is 12% less than at the start of that year.

(a) Find its value after 5 years, to the nearest dollar.

(b) Find the number of complete years after which its value first falls below $10 000.

(a) Each year the value is multiplied by $0.88$. After 5 years: $40\,000 \times 0.88^{5} = 21\,109.28\ldots$, so $21 109. ✓
(b) $40\,000 \times 0.88^{n} < 10\,000$, so $0.88^{n} < 0.25$. ✓
$n \lg 0.88 < \lg 0.25$. $\lg 0.88 < 0$, so $n > \dfrac{\lg 0.25}{\lg 0.88} = 10.84\ldots$ ✓
$n = 11$ years. Check: after 10 years about $11 140, after 11 years about $9803. ✓

Analysis

[5 marks]

A geometric progression has first term 90 and second term 60.

(a) Find the sum to infinity.

(b) Find the least value of $n$ for which the sum of the first $n$ terms exceeds 269.

(a) $r = \dfrac{60}{90} = \dfrac{2}{3}$, so $S_{\infty} = \dfrac{90}{1 - \frac{2}{3}} = 270$. ✓
(b) $S_{n} = \dfrac{90\left(1 - (\frac{2}{3})^{n}\right)}{\frac{1}{3}} = 270\left(1 - \left(\tfrac{2}{3}\right)^{n}\right) > 269$. ✓
$1 - \left(\tfrac{2}{3}\right)^{n} > \dfrac{269}{270}$, so $\left(\tfrac{2}{3}\right)^{n} < \dfrac{1}{270}$. ✓
$\lg \frac{2}{3} < 0$, so the inequality reverses: \[ n > \frac{\lg (1/270)}{\lg (2/3)} = 13.8\ldots \] ✓
$n = 14$. Check: $S_{13} \approx 268.61$, $S_{14} \approx 269.08$. ✓

Analysis

[5 marks]

A geometric progression is $1 + (2x - 1) + (2x - 1)^{2} + \dots$

(a) Find the set of values of $x$ for which the progression has a sum to infinity.

(b) Given that the sum to infinity is $\dfrac{4}{3}$, find $x$.

(a) $r = 2x - 1$. A sum to infinity needs $-1 < 2x - 1 < 1$. ✓
Add 1 and halve: $0 < x < 1$. ✓
(b) $\dfrac{1}{1 - (2x - 1)} = \dfrac{1}{2 - 2x} = \dfrac{4}{3}$, so $2 - 2x = \dfrac{3}{4}$ and $x = \dfrac{5}{8}$. ✓
$\dfrac{5}{8}$ lies in $0 < x < 1$, so it is valid ($r = \dfrac{1}{4}$). ✓
(c) $x = 1.5$ gives $r = 2$, so $|r| > 1$: each term is double the last and the sums grow without limit. ✓

Analysis

[6 marks]

An arithmetic progression has first term 12 and common difference $d$, where $d \ne 0$. The 1st, 3rd and 7th terms of the arithmetic progression are the first three terms of a geometric progression.

(a) Show that $d = 6$.

(b) Find the common ratio of the geometric progression.

(c) Find the sum of the first 8 terms of the geometric progression and the sum of the first 20 terms of the arithmetic progression.

(a) The AP terms are $12$, $12 + 2d$ and $12 + 6d$. For a GP, $(12 + 2d)^{2} = 12(12 + 6d)$. ✓
$144 + 48d + 4d^{2} = 144 + 72d$, so $4d^{2} = 24d$. ✓
$d \ne 0$, so divide by $4d$: $d = 6$. ✓
(b) The terms are $12, 24, 48$, so $r = 2$. ✓
(c) GP: $S_{8} = \dfrac{12(2^{8} - 1)}{2 - 1} = 12 \times 255 = 3060$. ✓
AP: $S_{20} = \dfrac{20}{2}\left(2 \times 12 + 19 \times 6\right) = 10 \times 138 = 1380$. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.

Find \(a\) and \(r\), then choose the formula.