Geometric series

In a nutshell

An geometric series finds the sum of a given number of terms in an geometric sequence, a sequence whose terms are separated by a common ratio.

Calculating the sum of a sequence

To find the sum of a given number of terms of an geometric sequence, use a formula rather than manually adding up the terms. The sum $S_n$ is given using the first term, $a$ , the common ratio, $r$ , and the $n^{th}$ term.

$\boxed {S_n = \dfrac {a(1-r^n)}{1-r} \quad or \quad S_n = \dfrac {a(r^n-1)}{r-1}}$

Note: Either formula can be used to find the sum but the formula on the left is easier when $r<1$ and the right when $r>1$ .

Deriving the formula

The terms in a finite geometric sequence are as follows:

$a, ar^1 ,ar^2,... , ar^{n-2}, ar^{n-1}$

This means the sum of a finite geometric sequence is equal to:

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

Multiplying the series by $r$ will give:

$\begin{aligned} rS_n &=&r &[a+ar^1 +...+ ar^{n-2}+ ar^{n-1}] \\ rS_n &=& &[ar+ar^2 +...+ ar^{n-1}+ ar^{n}] \end{aligned}$

Subtract this from $S_n$ :

$S_n - rS_n = [a+ar^1 +...+ ar^{n-2}+ ar^{n-1}]- [ar+ar^2 +...+ ar^{n-1}+ ar^{n}]$

$S_n(1-r ) = a+[ar^1-ar^1] +...+ [ar^{n-2}-ar^{n-2}]+ [ar^{n-1}-ar^{n-1}]- ar^{n}$

$S_n(1-r ) = a- ar^{n}$

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

Multiplying the top and bottom of the fraction by $-1$ which will maintain equivalence to $S_n$ :

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

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

Example 1

Find the sum of the first $12$ terms of the geometric sequence with a first term of, $7$ , and a common ratio of $0.9$ .

Identify the formula to use:

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

Substitute values to find the sum:

$S_{12} = \dfrac{7(1-0.9^{12})}{1-0.9}$

$\underline{S_{12} = 50.23\ (2 \ d.p.)}$

Example 2

How many terms are needed for the sum of a geometric series in the form $2+4+8+16 +...$ to exceed $1000$ ?

Identify the common ratio and first term:

$r=\dfrac 42 = 2 \\ a = 2$

Substitute the given values and solve:

$1000 = \dfrac{2(2^n-1)}{2- 1}$

$500= {2^n-1}$

$501 = 2^n$

$\ln (501) = n \ln(2)$

$n = \dfrac{\ln(501)}{\ln(2)} = 8.968666$

Round up since it is the sum which exceeds $1000$ :

$\underline{n = 9}$