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iGCSE (2021 Edition)

9.05 Geometric series


If the first term of a geometric sequence is $u_1$u1 and the common ratio is $r$r, then the sequence is given by:


Summing $n$n terms in a geometric sequence

Suppose we wish to add the first $n$n terms of this sequence. This will form a geometric series. We could write the sum as:


We saw a nifty trick for finding the formula for an arithmetic series by adding two sums together and pairing up terms. We will use a similar method here. If we multiply both sides of our geometric sum by the common ratio $r$r we see that:


Then, by carefully subtracting $rS_n$rSn from $S_n$Sn term by term, we see that all of the middle terms disappear:


This means that:


and when common factors are taken out on both sides of this equation, we find:


Finally, by dividing both sides by $\left(1-r\right)$(1r) (excluding the trivial case of $r=1$r=1) we reveal the geometric sum formula:


An extra step, multiplying the numerator and denominator by $-1$1, reveals a different form for $S_n$Sn. Both formulas will work in any situation, particularly when using a calculator. This form is generally easier to manage when the common ratio is greater than $r=1$r=1:


Geometric series

For any geometric sequence with starting value $u_1$u1 and common ratio $r$r, we can find the sum of the first $n$n terms, using:

$S_n=\frac{u_1\left(1-r^n\right)}{1-r}$Sn=u1(1rn)1r, for $r<1$r<1 or $S_n=\frac{u_1\left(r^n-1\right)}{r-1}$Sn=u1(rn1)r1, convenient if $r>1$r>1

The formulas for $S_n$Sn exclude the case for $r=1$r=1. For the case where $r=1$r=1, then the sequence becomes $u_1,u_1,u_1,u_1,\ldots$u1,u1,u1,u1,.

Is this a geometric sequence with $r=1$r=1 or an arithmetic sequence with $d=0$d=0? Either way, every term is clearly identical. Hence, the sum of the first $n$n terms is:

$S_n$Sn $=$= $u_1+u_1+u_1+...+u_1$u1+u1+u1+...+u1 ($n$n times)
$S_n$Sn $=$= $nu_1$nu1  

Worked example

Example 1

If the sum for the first $n$n terms of the geometric sequence $5,10,20,\ldots$5,10,20, is $5115$5115, find $n$n.

Think: We have an increasing geometric sequence. State $u_1$u1, $r$r and $S_n$Sn, then substitute into the formula $S_n=\frac{u_1\left(r^n-1\right)}{r-1}$Sn=u1(rn1)r1 and rearrange.

Do: $u_1=5$u1=5, $r=2$r=2 and $S_n=5115$Sn=5115, so we have:

$S_n$Sn $=$= $\frac{u_1\left(r^n-1\right)}{r-1}$u1(rn1)r1  
$5115$5115 $=$= $\frac{5\left(2^n-1\right)}{2-1}$5(2n1)21

Substitute values into formula


Hence,$5\left(2^n-1\right)$5(2n1) $=$= $5115$5115

Simplify fraction and bring unknown to left-hand side


$2^n-1$2n1 $=$= $1023$1023

Divide both sides by $5$5


$2^n$2n $=$= $1024$1024

Add $1$1 to both sides


$\therefore n$n $=$= $10$10

Solve for $n$n, using guess and check, technology or logarithms.


Practice questions

Question 1

Consider the series $4+12+36$4+12+36 ...

Find the sum of the first $12$12 terms.

Question 2

Consider the series $5+\frac{5}{2}+\frac{5}{4}$5+52+54 ...

  1. Find the common ratio, $r$r.

  2. Find the sum of the first $9$9 terms, rounding your answer to 1 decimal place.

Question 3

Consider the series $4-12+36-\text{. . . }-708588$412+36. . . 708588.

  1. Solve for $n$n, the number of terms in the series.

  2. Find the sum of the series.

Question 4

Average annual salaries are expected to increase by $4$4 percent each year. If the average annual salary this year is found to be $\$40000$$40000:

  1. Calculate the expected average annual salary in $6$6 years, correct to the nearest cent.

  2. This year, Vincent starts at a new job in which he will receive the average annual salary for each year of his employment. Over the coming $6$6 years (including this year) he plans to save half of each year’s annual salary.

    What will be his total savings over these $6$6 years? Give your answer correct to the nearest cent.



Recognise geometric progressions.


Use the formulae for the nth term and for the sum of the first n terms to solve problems involving geometric progressions.

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