Geometric Sequences and Series

Geometric Sequences and Series — Moosa Academy

Replace the constant addition of an arithmetic sequence with a constant multiplication and the behaviour changes completely. Terms no longer creep along a straight line; they race away, collapse toward zero, or flip sign at every step — all decided by a single number, the common ratio.

Theorem The two core formulas

The $n$ th term is $a_n = a_1 \cdot r^{\,n-1}$ , and the sum of the first $n$ terms is $S_n = \dfrac{a_1(1 - r^{\,n})}{1 - r}$ , for $r \neq 1$ .

Here $a_1$ is the first term, $r$ the common ratio, and $n$ the position.

Term formula:

a_n = a_1 \cdot r^{\,n-1}

Sum, knowing $r$ and $n$ :

S_n = \dfrac{a_1(1 - r^{\,n})}{1 - r}

Sum, knowing the last term:

S_n = \dfrac{a_1 - a_n r}{1 - r}

Find $r$ by dividing any term by the one before it, just as you found $d$ by subtracting.

Concept What the ratio controls

r > 1

— the terms grow

0 < r < 1

— the terms shrink toward zero

r < 0

— the terms alternate in sign

Each bar is double the one before it. After only a handful of steps the growth is dramatic — which is exactly why chain letters and compound interest are modelled this way.

Example Finding a term

a) An email chain

Ahmed sends $5$ emails in the first stage, and every recipient forwards it to $5$ more. How many emails are sent in the eighth stage?

a_1 = 5

r = 5

n = 8

Substitute:

a_8 = 5 \cdot 5^{\,7}

Evaluate:

5^{\,7} = 78125

$a_8 = 390625$ emails

b) Write a rule for $0.5,\ 2,\ 8,\ 32,\ \ldots$

Common ratio:

2 \div 0.5 = 4

$a_n = 0.5 \cdot 4^{\,n-1}$

c) Given $a_4 = 5$ and $r = 6$ , find $a_1$

Substitute:

5 = a_1 \cdot 6^{\,3}

Simplify:

5 = 216\,a_1

$a_1 = \dfrac{5}{216}$

Example Geometric means

Find three geometric means between $2$ and $1250$ .

Three means plus the two end terms gives

n = 5

Substitute:

1250 = 2 \cdot r^{\,4}

Simplify:

r^{\,4} = 625

Take the fourth root:

r = 5

r = -5

r = 5

2,\ 10,\ 50,\ 250,\ 1250

r = -5

2,\ -10,\ 50,\ -250,\ 1250

The means are $10,\ 50,\ 250$ or $-10,\ 50,\ -250$

An even power hides a sign, so both answers are genuine. This never happens with arithmetic means, where solving for $d$ is a linear step with a single result.

Example Summing a series

a) Total emails through the eighth stage

a_1 = 5

r = 5

n = 8

Substitute:

S_8 = \dfrac{5(1 - 5^{\,8})}{1 - 5}

Evaluate the power:

5^{\,8} = 390625

Simplify:

S_8 = \dfrac{5(-390624)}{-4}

$S_8 = 488280$ emails in total

b) Evaluate $\displaystyle\sum_{k=3}^{8} 3 \cdot 2^{\,k}$

First term at

k = 3

3 \cdot 2^{\,3} = 24

Number of terms:

8 - 3 + 1 = 6

Common ratio:

r = 2

Substitute:

S_6 = \dfrac{24(1 - 2^{\,6})}{1 - 2} = \dfrac{24(-63)}{-1}

$S_6 = 1512$

Counting the terms is where marks are lost. From $k = 3$ to $k = 8$ there are $6$ terms, not $5$ — subtract, then add one back.

Summary

A geometric sequence multiplies by a constant ratio $r$ instead of adding a constant $d$ .
Find $r$ by dividing any term by the one before it.
The $n$ th term is $a_n = a_1 \cdot r^{\,n-1}$ .
The sum of $n$ terms is $S_n = \dfrac{a_1(1 - r^{\,n})}{1 - r}$ , valid whenever $r \neq 1$ .
$r > 1$ grows, $0 < r < 1$ shrinks, and $r < 0$ alternates in sign.
Inserting $m$ geometric means makes $n = m + 2$ terms, and an even root gives two valid answers.
In sigma notation the number of terms is top $-$ bottom $+ 1$ .