Geometric Sequences and Series – A Level Maths Year 2

1 Geometric Sequences

A geometric sequence (or geometric progression, GP) is multiplied by the same number every time. That number is the common ratio $r$, found by dividing any term by the one before it.

The $n$th term of a geometric sequence

$u_n = ar^{n-1}$

Sequence	$a$	$r$	Behaviour
$3, 6, 12, 24, \ldots$	$3$	$2$	Grows without limit
$80, 40, 20, 10, \ldots$	$80$	$\tfrac12$	Decreases towards $0$
$2, -6, 18, -54, \ldots$	$2$	$-3$	Alternates in sign, growing in size
$5, 5, 5, \ldots$	$5$	$1$	Constant

Worked Example 1 — The $n$th term

Find the $10$th term of $3, 6, 12, \ldots$

①$a = 3$, $r = 6 \div 3 = 2$.

②$u_{10} = 3 \times 2^{9} = 3 \times 512 = 1536$.

The power is $n - 1 = 9$, not $10$. The first term has $r^0$.

Worked Example 2 — Two terms given

The $3$rd term of a geometric sequence is $18$ and the $6$th term is $486$. Find $a$ and $r$.

①$ar^2 = 18$ and $ar^5 = 486$.

②Divide: $r^3 = 27$, so $r = 3$.

③$a = 18 \div 9 = 2$.

With a geometric sequence you divide the equations to eliminate $a$. With an arithmetic one you subtract.

Worked Example 3 — Three consecutive terms

$x$, $x + 8$ and $9x$ are consecutive terms of a geometric sequence. Find the possible values of $x$.

①The ratios are equal: $\dfrac{x+8}{x} = \dfrac{9x}{x+8}$, so $(x + 8)^2 = 9x^2$.

②$x + 8 = \pm 3x$.

③$x + 8 = 3x \Rightarrow x = 4$, giving $4, 12, 36$ with $r = 3$.

④$x + 8 = -3x \Rightarrow x = -2$, giving $-2, 6, -18$ with $r = -3$.

2 Geometric Series and the Sum Formula

Sum of the first $n$ terms of a geometric series ($r \neq 1$)

$S_n = \dfrac{a(1 - r^n)}{1 - r} \qquad\text{or equivalently}\qquad S_n = \dfrac{a(r^n - 1)}{r - 1}$

Both forms are identical (multiply top and bottom by $-1$). Use the second when $r > 1$ so that both brackets are positive.

Proof — The sum of a geometric series

①$S_n = a + ar + ar^2 + \cdots + ar^{n-1}$

②Multiply by $r$: $rS_n = ar + ar^2 + \cdots + ar^{n-1} + ar^n$

③Subtract: almost every term cancels, leaving $S_n - rS_n = a - ar^n$

④$S_n(1 - r) = a(1 - r^n)$

⑤$S_n = \dfrac{a(1 - r^n)}{1 - r}$ for $r \neq 1$. $\blacksquare$

Dividing by $1 - r$ is why the formula needs $r \neq 1$. When $r = 1$ every term is $a$ and $S_n = na$.

Worked Example 4 — Using the sum formula

Find the sum of the first $10$ terms of $5 + 10 + 20 + \cdots$

①$a = 5$, $r = 2$, $n = 10$.

②$S_{10} = \dfrac{5(2^{10} - 1)}{2 - 1} = 5 \times 1023 = 5115$.

Worked Example 5 — Sigma notation

Evaluate $\displaystyle\sum_{r=1}^{8} 3 \times 2^{r}$.

①The first term ($r = 1$) is $6$ and each term doubles: $a = 6$, ratio $2$, $8$ terms.

②$S_8 = \dfrac{6(2^8 - 1)}{2 - 1} = 6 \times 255 = 1530$.

Do not confuse the sigma variable $r$ with the common ratio. Here the ratio is $2$ and the first term is $3 \times 2^1$, not $3$.

3 Convergence and the Sum to Infinity

If $|r| < 1$ then $r^n \to 0$ as $n \to \infty$. The sum $S_n = \dfrac{a(1 - r^n)}{1 - r}$ therefore gets closer and closer to $\dfrac{a}{1-r}$. The series is convergent. If $|r| \geqslant 1$ the terms do not shrink and the series is divergent — there is no sum to infinity.

Sum to infinity — valid only when $|r| < 1$, i.e. $-1 < r < 1$

$S_\infty = \dfrac{a}{1 - r}$

Partial sums of $1 + \tfrac12 + \tfrac14 + \cdots$ approach, but never reach, $S_\infty = 2$.

Worked Example 6 — Finding a sum to infinity

Find the sum to infinity of $24 + 12 + 6 + 3 + \cdots$

①$a = 24$, $r = \tfrac12$. Since $|r| < 1$ the sum to infinity exists.

②$S_\infty = \dfrac{24}{1 - \frac12} = 48$.

Worked Example 7 — Working backwards

A geometric series has sum to infinity equal to four times its first term. Find the common ratio.

①$\dfrac{a}{1 - r} = 4a$.

②$a \neq 0$, so divide by $a$: $1 = 4(1 - r)$, giving $r = \tfrac34$.

③$|r| < 1$, so the sum to infinity does exist ✓

Worked Example 8 — A condition for convergence

The series $1 + 2x + 4x^2 + 8x^3 + \cdots$ is convergent. (a) Find the set of possible values of $x$. (b) Given that the sum to infinity is $3$, find $x$.

①(a) The common ratio is $2x$. Convergence needs $|2x| < 1$, so $-\tfrac12 < x < \tfrac12$.

②(b) $\dfrac{1}{1 - 2x} = 3 \Rightarrow 1 - 2x = \tfrac13 \Rightarrow x = \tfrac13$.

③$\tfrac13$ lies in the interval from (a) ✓

4 Using Logarithms to Find $n$

When $n$ is the unknown, it sits in a power. Rearrange to the form $r^n \gtrless k$ and take logarithms of both sides.

The sign trap. If $0 < r < 1$ then $\ln r$ is negative, so dividing by it reverses the inequality. For example $0.8^n < 0.1 \Rightarrow n \ln 0.8 < \ln 0.1 \Rightarrow n > \dfrac{\ln 0.1}{\ln 0.8}$.

Worked Example 9 — The least $n$ for a sum

Find the least value of $n$ for which the sum of the first $n$ terms of $3 + 3.3 + 3.63 + \cdots$ exceeds $100$.

①$a = 3$, $r = 1.1$: $S_n = \dfrac{3(1.1^n - 1)}{0.1} = 30(1.1^n - 1)$.

②$30(1.1^n - 1) > 100 \Rightarrow 1.1^n > \tfrac{13}{3}$.

③$n \ln 1.1 > \ln\tfrac{13}{3} \Rightarrow n > \dfrac{\ln(13/3)}{\ln 1.1} = 15.38\ldots$

④Least $n = 16$. Check: $S_{15} \approx 95.3$, $S_{16} \approx 107.8$ ✓

5 Modelling: Savings and Depreciation

A percentage change each period is a geometric model. An increase of $p\%$ gives $r = 1 + \tfrac{p}{100}$; a decrease of $p\%$ gives $r = 1 - \tfrac{p}{100}$.

Worked Example 10 — Depreciation

A car costs £$18\,000$ and loses $15\%$ of its value each year. (a) Find its value after $5$ years. (b) After how many complete years is it first worth less than £$5000$?

①Value after $n$ years: $V = 18\,000 \times 0.85^n$.

②(a) $18\,000 \times 0.85^5 = $ £$7986.70$ (nearest penny).

③(b) $18\,000 \times 0.85^n < 5000 \Rightarrow 0.85^n < \tfrac{5}{18}$.

④$n > \dfrac{\ln(5/18)}{\ln 0.85} = 7.88\ldots$ (inequality reversed, as $\ln 0.85 < 0$).

⑤After $8$ years. Check: $V_7 \approx $ £$5770$, $V_8 \approx $ £$4905$ ✓

Worked Example 11 — Regular savings

At the start of each year Leah pays £$500$ into an account paying $3\%$ compound interest per year. How much is in the account at the end of $10$ years?

①The first £$500$ earns interest for $10$ years: worth $500 \times 1.03^{10}$.

②The last £$500$ earns interest for $1$ year: worth $500 \times 1.03$.

③Total $= 500(1.03) + 500(1.03)^2 + \cdots + 500(1.03)^{10}$ — a GP with $a = 515$, $r = 1.03$, $n = 10$.

④$S_{10} = \dfrac{515(1.03^{10} - 1)}{0.03} = $ £$5903.90$ (nearest penny).

Write out the first and last terms in words before using the formula. Most errors in savings questions come from being one year out at one end.

6 Quick Reference

$n$th term

$u_n = ar^{n-1}$

Common ratio

$r = u_{n+1} \div u_n$

Sum

$S_n = \dfrac{a(1-r^n)}{1-r}$, $r \neq 1$

Proof

Write $S_n$ and $rS_n$, then subtract.

Sum to infinity

$S_\infty = \dfrac{a}{1-r}$ only if $|r| < 1$.

Consecutive terms

$(\text{middle})^2 = \text{first} \times \text{third}$

Two terms given

Divide to eliminate $a$.

Finding $n$

Take logs; reverse the inequality if dividing by $\ln r < 0$.

Percentages

$+p\%$: $r = 1 + \tfrac{p}{100}$; $-p\%$: $r = 1 - \tfrac{p}{100}$.

7 Practice Questions

Question 1

Find the $8$th term of $2, -6, 18, \ldots$

▶ Show solution

$a = 2$, $r = -3$: $u_8 = 2 \times (-3)^7 = 2 \times (-2187) = -4374$.

Question 2

Find the sum of the first $8$ terms of $1 + 3 + 9 + \cdots$

▶ Show solution

$S_8 = \dfrac{1(3^8 - 1)}{3 - 1} = \dfrac{6560}{2} = 3280$.

Question 3

Find the sum to infinity of $81 - 27 + 9 - 3 + \cdots$

▶ Show solution

$a = 81$, $r = -\tfrac13$, and $|r| < 1$.

$S_\infty = \dfrac{81}{1 + \frac13} = \dfrac{81}{\frac43} = \dfrac{243}{4} = 60.75$.

Question 4

The $2$nd term of a geometric sequence is $12$ and the $5$th term is $324$. Find the sum of the first $6$ terms.

▶ Show solution

$ar = 12$, $ar^4 = 324$. Divide: $r^3 = 27$, $r = 3$, $a = 4$.

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

Question 5

Evaluate $\displaystyle\sum_{r=1}^{10} 3(2^r)$.

▶ Show solution

First term $6$, ratio $2$, $10$ terms.

$S_{10} = \dfrac{6(2^{10} - 1)}{1} = 6 \times 1023 = 6138$.

Question 6

A geometric series has second term $6$ and sum to infinity $27$. Find the two possible values of the common ratio and the corresponding first terms.

▶ Show solution

$ar = 6$ and $\dfrac{a}{1-r} = 27$, so $a = 27(1 - r)$.

Substitute: $27r(1 - r) = 6 \Rightarrow 27r^2 - 27r + 6 = 0 \Rightarrow 9r^2 - 9r + 2 = 0$.

$(3r - 1)(3r - 2) = 0$, so $r = \tfrac13$ or $r = \tfrac23$ (both satisfy $|r| < 1$).

$r = \tfrac13$: $a = 18$. $r = \tfrac23$: $a = 9$.

Question 7

The series $1 + \dfrac{x-1}{3} + \left(\dfrac{x-1}{3}\right)^2 + \cdots$ is convergent. Find the range of possible values of $x$, and the sum to infinity when $x = 2$.

▶ Show solution

Need $\left|\dfrac{x-1}{3}\right| < 1$, so $-3 < x - 1 < 3$, giving $-2 < x < 4$.

When $x = 2$, $r = \tfrac13$: $S_\infty = \dfrac{1}{1 - \frac13} = \dfrac32$.

Question 8

A geometric series has first term $a$ and common ratio $0.8$. Find the least $n$ such that $S_n$ is within $0.1\%$ of $S_\infty$, i.e. $S_n > 0.999\,S_\infty$.

▶ Show solution

$\dfrac{S_n}{S_\infty} = \dfrac{a(1 - 0.8^n)/0.2}{a/0.2} = 1 - 0.8^n$.

$1 - 0.8^n > 0.999 \Rightarrow 0.8^n < 0.001$.

$n > \dfrac{\ln 0.001}{\ln 0.8} = 30.96\ldots$ (reversed since $\ln 0.8 < 0$), so $n = 31$.

Question 9

A ball is dropped from a height of $10$ m. After each bounce it rises to $\tfrac34$ of the height it fell from. (a) Find the height it reaches after the $5$th bounce. (b) Find the total distance it travels before coming to rest, assuming it bounces indefinitely.

▶ Show solution

(a) $10 \times 0.75^5 = 2.37$ m (3 s.f.).

(b) It falls $10$ m, then each bounce goes up and down the same distance.

The rebound heights $7.5, 5.625, \ldots$ form a GP with $a = 7.5$, $r = 0.75$, so their sum is $\dfrac{7.5}{0.25} = 30$ m.

Total $= 10 + 2 \times 30 = 70$ m.

Question 10

Ali saves £$100$ in the first month, and each month saves $2\%$ more than the month before.

(a) How much does Ali save in month $12$? (b) Find his total savings over the first $12$ months. (c) Show that the total after $n$ months exceeds £$5000$ when $1.02^n > 2$, and hence find the first month in which the total exceeds £$5000$. (d) Give one reason why the model may not be realistic in the long term.

▶ Show solution

(a) $u_{12} = 100 \times 1.02^{11} = $ £$124.34$.

(b) $S_{12} = \dfrac{100(1.02^{12} - 1)}{0.02} = $ £$1341.21$.

(c) $\dfrac{100(1.02^n - 1)}{0.02} > 5000 \Rightarrow 5000(1.02^n - 1) > 5000 \Rightarrow 1.02^n > 2$ ✓

$n > \dfrac{\ln 2}{\ln 1.02} = 35.003\ldots$, so $n = 36$.

Check: $S_{35} \approx $ £$4999.45$ — just short — and $S_{36} \approx $ £$5199.44$ ✓. The first month is month $36$.

(d) The monthly amount grows exponentially without limit, but Ali's income is unlikely to grow at $2\%$ a month for ever.

Geometric Sequences and Series · A Level Maths Year 2 · Created with MathJax