Geometric progressions

AS · P1 · 14 min

A geometric progression (GP) is a list of numbers where you multiply by the same amount each step, like $3, 6, 12, 24, \ldots$ or $80, 40, 20, 10, \ldots$ . Anything that grows or shrinks by a fixed percentage is a GP: compound interest, depreciation, a population growing by $3\%$ a year. Paper 1 asks you to find terms and sums, to find the first term and ratio from given facts, and to decide how many terms are needed to pass a target, all without logarithms.

What makes a progression geometric

In an arithmetic progression consecutive terms have the same difference. In a geometric progression they have the same ratio.

Definition

A geometric progression is a sequence in which each term is obtained from the previous one by multiplying by a fixed non-zero number $r$ , called the common ratio. The first term is written $a$ .

So $\dfrac{u_{n+1}}{u_n} = r$ for every $n$ .

To find $r$ , divide a term by the one before it:

Sequence	$a$	$r$	Behaviour
$3, 6, 12, 24, \ldots$	$3$	$2$	grows
$80, 40, 20, 10, \ldots$	$80$	$\tfrac{1}{2}$	shrinks towards $0$
$48, -24, 12, -6, \ldots$	$48$	$-\tfrac{1}{2}$	alternates in sign, shrinks
$2, -6, 18, -54, \ldots$	$2$	$-3$	alternates in sign, grows

A negative ratio makes the signs alternate. A ratio between $-1$ and $1$ makes the terms shrink in size; a ratio bigger than $1$ in size makes them grow.

The nth term

u_1 = a, \quad u_2 = ar, \quad u_3 = ar^2, \quad u_4 = ar^3, \quad \ldots

As with an AP, reaching the $n$ th term takes $n - 1$ steps, so the power is $n - 1$ .

Key result

The $n$ th term of a GP with first term $a$ and common ratio $r$ is

u_n = ar^{n-1}

Three non-zero numbers $p$ , $q$ , $r$ are consecutive terms of a GP exactly when $\dfrac{q}{p} = \dfrac{r}{q}$ , that is

q^2 = pr

y = 0.75 * 2^x (1, 1.5) (2, 3) (3, 6) (4, 12) (5, 24) y = 32 * 0.5^x (1, 16) (2, 8) (3, 4) (4, 2) (5, 1) (6, 0.5)

The terms of $1.5, 3, 6, 12, \ldots$ ( $r = 2$ ) and $16, 8, 4, 2, \ldots$ ( $r = \tfrac{1}{2}$ ) plotted against $n$ . A GP's terms lie on an exponential curve, not a line: equal steps in $n$ multiply the term by the same factor.

Finding the ratio from two terms

If two terms are given, write both in the form $ar^{n-1}$ and divide the equations. The $a$ cancels and leaves a power of $r$ . For example, $u_2 = 12$ and $u_5 = 324$ give

\frac{ar^4}{ar} = \frac{324}{12} \quad\Rightarrow\quad r^3 = 27 \quad\Rightarrow\quad r = 3

If the power left is even, such as $r^2 = 9$ , there are two possible ratios, $r = 3$ and $r = -3$ . Keep both unless the question rules one out (for example "all the terms are positive").

The sum of the first n terms

The subtraction trick

Write out the sum, then multiply every term by $r$ :

\begin{aligned} S_n &= a + ar + ar^2 + \cdots + ar^{n-1} \\ rS_n &= \phantom{a + {}} ar + ar^2 + \cdots + ar^{n-1} + ar^n \end{aligned}

Every term except the first of the top line and the last of the bottom line appears in both. Subtracting:

S_n - rS_n = a - ar^n \quad\Rightarrow\quad S_n(1 - r) = a(1 - r^n) \quad\Rightarrow\quad S_n = \frac{a(1 - r^n)}{1 - r}

This needs $r \neq 1$ . (If $r = 1$ every term is $a$ and $S_n = na$ .)

Key result

The sum of the first $n$ terms of a GP is

S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a(r^n - 1)}{r - 1} \qquad (r \neq 1)

The two forms are identical (multiply top and bottom by $-1$ ). Both appear in MF19 as the first; use the $r^n - 1$ form when $r > 1$ so that the numbers stay positive.

When $-1 < r < 1$ the power $r^n$ gets smaller and smaller, so $S_n$ settles down towards $\dfrac{a}{1 - r}$ . That limit is the sum to infinity, which has its own note: Sum to infinity.

Solving a GP problem

Translate each fact into an equation using $u_n = ar^{n-1}$ or $S_n = \dfrac{a(1 - r^n)}{1 - r}$ .
To find $r$ , divide one equation by another so that $a$ cancels. For sums, factorise first: $1 - r^4 = (1 - r^2)(1 + r^2)$ .
Check how many values of $r$ are possible and which ones the question allows.
Substitute back to find $a$ .
For "least $n$ " questions, reduce to $r^n > k$ (or $r^n < k$ ) and find $n$ by trying consecutive integers on your calculator, writing down the two values either side.

Finding n without logarithms

Logarithms are not part of Paper 1. When you need the least $n$ with $1.5^n > 126$ , use your calculator to evaluate powers until you pass the target, and show the two that bracket it:

1.5^{11} \approx 86.5 < 126, \qquad 1.5^{12} \approx 129.7 > 126

so $n = 12$ . Writing these two values earns the method mark.

Looking ahead

In Paper 3 you will solve $1.5^n = 126$ directly as $n = \dfrac{\ln 126}{\ln 1.5}$ . See Logarithms. On Paper 1, trial is expected and fully acceptable.

Worked examples

First term and ratio from two terms

The second term of a geometric progression is $12$ and the fifth term is $324$ . Find the common ratio, the first term and the sum of the first $6$ terms.

Solution

ar = 12, \qquad ar^4 = 324

Dividing, $r^3 = 27$ , so $r = 3$ . Then $a = \dfrac{12}{3} = 4$ .

S_6 = \frac{4(3^6 - 1)}{3 - 1} = \frac{4 \times 728}{2} = 1456

A negative common ratio

For the geometric progression $48, -24, 12, \ldots$ find the eighth term and the sum of the first eight terms, giving exact answers.

Solution

$r = \dfrac{-24}{48} = -\dfrac{1}{2}$ .

u_8 = 48\left(-\tfrac{1}{2}\right)^7 = -\frac{48}{128} = -\frac{3}{8}

S_8 = \frac{48\left(1 - \left(-\tfrac{1}{2}\right)^8\right)}{1 - \left(-\tfrac{1}{2}\right)} = \frac{48\left(1 - \tfrac{1}{256}\right)}{\tfrac{3}{2}} = 32 \times \frac{255}{256} = \frac{255}{8}

Brackets around $-\tfrac{1}{2}$ matter: $\left(-\tfrac12\right)^8$ is positive, and $1 - r = 1 + \tfrac{1}{2}$ .

Three terms in a geometric progression

The first three terms of a geometric progression are $2k + 3$ , $k + 6$ and $k$ , where $k$ is a positive constant.

(a) Find the value of $k$ .

(b) Find the common ratio and the fourth term.

Solution

(a) Consecutive terms of a GP satisfy $(\text{middle})^2 = \text{first} \times \text{third}$ :

\begin{aligned} (k + 6)^2 &= k(2k + 3) \\ k^2 + 12k + 36 &= 2k^2 + 3k \\ 0 &= k^2 - 9k - 36 \\ 0 &= (k - 12)(k + 3) \end{aligned}

Since $k$ is positive, $k = 12$ .

(b) The terms are $27, 18, 12$ , so $r = \dfrac{18}{27} = \dfrac{2}{3}$ and $u_4 = 12 \times \dfrac{2}{3} = 8$ .

Using a ratio of sums

All the terms of a geometric progression are positive. The sum of the first two terms is $8$ and the sum of the first four terms is $80$ . Find the first term and the common ratio.

Solution

Write the sums out directly:

\begin{aligned} a + ar &= a(1 + r) = 8 \\ a + ar + ar^2 + ar^3 &= a(1 + r)(1 + r^2) = 80 \end{aligned}

(The second factorises because $a + ar + ar^2 + ar^3 = a(1 + r) + ar^2(1 + r)$ .) Dividing,

1 + r^2 = 10 \quad\Rightarrow\quad r^2 = 9 \quad\Rightarrow\quad r = \pm 3

All the terms are positive, so $r = 3$ , and $a = \dfrac{8}{1 + 3} = 2$ . Check: $2 + 6 + 18 + 54 = 80$ .

Least number of terms

A geometric progression has first term $2$ and common ratio $1.5$ . Find the least number of terms for which the sum exceeds $500$ .

Solution

S_n = \frac{2(1.5^n - 1)}{1.5 - 1} = 4(1.5^n - 1)

We need $4(1.5^n - 1) > 500$ , so $1.5^n - 1 > 125$ , that is $1.5^n > 126$ .

1.5^{11} \approx 86.5, \qquad 1.5^{12} \approx 129.7

So the least number of terms is $n = 12$ .

Depreciation

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

(a) Find the value of the machine at the end of $6$ years.

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

Solution

Losing $12\%$ means keeping $88\%$ , so the values are multiplied by $r = 0.88$ each year.

(a) After $6$ years the value is

40\,000 \times 0.88^6 \approx 18\,576 \text{ dollars} \quad (18\,600 \text{ to 3 s.f.})

(b) We need $40\,000 \times 0.88^n < 10\,000$ , that is $0.88^n < 0.25$ .

0.88^{10} \approx 0.2785 > 0.25, \qquad 0.88^{11} \approx 0.2451 < 0.25

The value first falls below $10\,000$ dollars after $11$ years.

A salary scheme

An employee's starting salary is $30\,000$ dollars a year. Each year the salary increases by $5\%$ of the previous year's salary.

(a) Find the total amount earned in the first $10$ years, to the nearest dollar.

(b) In which year does the salary first exceed $45\,000$ dollars?

Solution

The annual salaries form a GP with $a = 30\,000$ and $r = 1.05$ .

(a)

S_{10} = \frac{30\,000(1.05^{10} - 1)}{1.05 - 1} = 600\,000(1.05^{10} - 1) \approx 377\,337 \text{ dollars}

(b) The salary in year $n$ is $30\,000 \times 1.05^{n-1}$ . We need $1.05^{n-1} > 1.5$ .

1.05^{8} \approx 1.477, \qquad 1.05^{9} \approx 1.551

So $n - 1 = 9$ and the salary first exceeds $45\,000$ dollars in the $10$ th year. Notice the power is $n - 1$ , not $n$ : the first year's salary is $30\,000 \times 1.05^0$ .

Watch out

Wrong power. The $n$ th term is $ar^{n-1}$ , so $u_8 = ar^7$ . In context, the salary in year $n$ is $30\,000 \times 1.05^{n-1}$ , but the value of the machine "after $n$ years" is $40\,000 \times 0.88^n$ , because the starting value is the value after $0$ years. Decide what the first term represents before writing the power.

Percentage changes. An increase of $5\%$ means $r = 1.05$ , not $r = 0.05$ . A decrease of $12\%$ means $r = 0.88$ , not $r = -0.12$ or $r = 1.12$ .

Losing a root. $r^2 = 9$ gives $r = 3$ or $r = -3$ . Only reject one with a stated reason.

Brackets with negative ratios. $-\tfrac{1}{2}^8$ on a calculator means $-(\tfrac12)^8$ . Type $\left(-\tfrac{1}{2}\right)^8$ .

Adding instead of multiplying. In a GP, $u_3 = u_1 r^2$ , not $u_1 + 2r$ . If a question says "increases by the same amount" it is an AP; "increases by the same percentage" or "by a factor" is a GP.

Exam tip

Equations first. Write $ar = 12$ and $ar^4 = 324$ before solving. A typical 3-mark part gives one mark for a correct pair of equations, one for eliminating $a$ , and one for the answers.
Least $n$ by trial. Show the two powers either side of the target, as in the examples. Writing only " $n = 12$ " risks losing method marks.
Rounding in context. Keep full calculator values until the end, then round money sensibly (to the nearest dollar, or 3 s.f.). Rounding $r^n$ early can change the answer to a "least $n$ " question.
Exact answers. If the question says "exact", leave fractions such as $\tfrac{255}{8}$ ; do not convert to decimals.
AP or GP? Many questions put the two side by side and ask you to compare them. See Problems combining progressions.

Summary

A GP multiplies by a constant ratio $r$ ; find it as a term divided by the term before.
$u_n = ar^{n-1}$ .
$S_n = \dfrac{a(1 - r^n)}{1 - r} = \dfrac{a(r^n - 1)}{r - 1}$ for $r \neq 1$ , proved by subtracting $rS_n$ from $S_n$ .
$p, q, r$ are consecutive terms of a GP exactly when $q^2 = pr$ .
To find $r$ , divide equations so $a$ cancels; an even power gives two possible ratios.
Percentage growth of $p\%$ gives $r = 1 + \tfrac{p}{100}$ ; a decrease gives $r = 1 - \tfrac{p}{100}$ .
On Paper 1 find the least $n$ by trial, showing the values either side.
When $|r| < 1$ the sums approach the sum to infinity $\dfrac{a}{1 - r}$ .

Practice questions

Question

Find the $8$ th term and the sum of the first $8$ terms of the geometric progression $81, 54, 36, \ldots$ , giving exact answers.
The numbers $x$ , $x + 6$ and $x + 15$ are consecutive terms of a geometric progression. Find $x$ and the common ratio.
The third term of a GP is $\tfrac{9}{4}$ and the sixth term is $\tfrac{243}{32}$ . Find the first term, the common ratio and the sum of the first six terms.
A GP has first term $5$ and common ratio $-2$ . Find the sum of the first $10$ terms.
Find the least number of terms of the GP $3, 6, 12, \ldots$ needed for the sum to exceed $1000$ .
The second term of a GP is $6$ and the fourth term is $54$ . Given that the common ratio is negative, find the first term and the sum of the first five terms.
The population of a town is $2000$ and increases by $3\%$ each year. Find the number of complete years after which the population first exceeds $3000$ .
Find the sum of the $6$ th to the $10$ th terms inclusive of the GP $2, 6, 18, \ldots$
The sum of the first three terms of a GP is seven times the first term. (a) Find the two possible values of the common ratio. (b) Given that the common ratio is positive and the third term is $36$ , find the least value of $n$ for which $S_n > 10\,000$ .
A geometric progression has common ratio $r$ , where $r \neq 1$ , and $S_n$ denotes the sum of the first $n$ terms. (a) Show that $S_{2n} = S_n(1 + r^n)$ . (b) Given that $S_8 = 17S_4$ , find the possible values of $r$ . (c) Given also that the second term is $6$ and $r > 0$ , find $S_8$ .

Answers

$r = \tfrac{54}{81} = \tfrac{2}{3}$ . $u_8 = 81\left(\tfrac{2}{3}\right)^7 = \dfrac{81 \times 128}{2187} = \dfrac{128}{27}$ . $S_8 = \dfrac{81\left(1 - \left(\tfrac{2}{3}\right)^8\right)}{\tfrac{1}{3}} = 243\left(1 - \tfrac{256}{6561}\right) = \dfrac{6305}{27}$ .
$(x + 6)^2 = x(x + 15)$ , so $x^2 + 12x + 36 = x^2 + 15x$ , giving $3x = 36$ and $x = 12$ . The terms are $12, 18, 27$ , so $r = \tfrac{3}{2}$ .
$ar^2 = \tfrac{9}{4}$ and $ar^5 = \tfrac{243}{32}$ . Dividing, $r^3 = \tfrac{243}{32} \times \tfrac{4}{9} = \tfrac{27}{8}$ , so $r = \tfrac{3}{2}$ . Then $a = \tfrac{9}{4} \div \tfrac{9}{4} = 1$ . $S_6 = \dfrac{1.5^6 - 1}{0.5} = 2(11.390625 - 1) = \dfrac{665}{32} = 20.78125$ .
$S_{10} = \dfrac{5\left(1 - (-2)^{10}\right)}{1 - (-2)} = \dfrac{5(1 - 1024)}{3} = -1705$ .
$S_n = \dfrac{3(2^n - 1)}{2 - 1} = 3(2^n - 1) > 1000$ , so $2^n > 334.3$ . $2^8 = 256$ , $2^9 = 512$ , so $n = 9$ .
$ar = 6$ and $ar^3 = 54$ , so $r^2 = 9$ and $r = -3$ (negative). Then $a = -2$ . $S_5 = \dfrac{-2\left(1 - (-3)^5\right)}{1 - (-3)} = \dfrac{-2(1 + 243)}{4} = -122$ . (Terms: $-2, 6, -18, 54, -162$ .)
Population after $n$ years is $2000 \times 1.03^n$ . Need $1.03^n > 1.5$ . $1.03^{13} \approx 1.469$ and $1.03^{14} \approx 1.513$ , so after $14$ years.
$a = 2$ , $r = 3$ . Sum of terms $6$ to $10$ is $S_{10} - S_5 = (3^{10} - 1) - (3^5 - 1) = 59\,049 - 243 = 58\,806$ .
(a) $a + ar + ar^2 = 7a$ . Dividing by $a$ (non-zero), $r^2 + r - 6 = 0$ , so $(r + 3)(r - 2) = 0$ and $r = 2$ or $r = -3$ . (b) $r = 2$ and $ar^2 = 4a = 36$ , so $a = 9$ . $S_n = 9(2^n - 1) > 10\,000$ gives $2^n > 1112.1$ . $2^{10} = 1024$ , $2^{11} = 2048$ , so $n = 11$ .
(a) $S_{2n} = \dfrac{a(1 - r^{2n})}{1 - r} = \dfrac{a(1 - r^n)(1 + r^n)}{1 - r} = S_n(1 + r^n)$ , using the difference of two squares $1 - r^{2n} = (1 - r^n)(1 + r^n)$ . (b) With $n = 4$ : $S_8 = S_4(1 + r^4) = 17S_4$ . Since $S_4 \neq 0$ , $1 + r^4 = 17$ , so $r^4 = 16$ and $r = 2$ or $r = -2$ . (c) $r = 2$ and $ar = 6$ give $a = 3$ . $S_8 = 3(2^8 - 1) = 765$ . (Check: $S_4 = 45$ and $17 \times 45 = 765$ .)

What makes a progression geometric

The nth term

Finding the ratio from two terms

The sum of the first n terms

The subtraction trick

Finding n without logarithms

Worked examples

Practice questions

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