Deriving the Sum of the Geometric Sequence

In the previous posts, you have learned about arithmetic sequence and we have discussed how to find its sum. In this post, we discuss another type of sequence, the geometric sequence or geometric progression.

A sequence is said to be in geometric progression if each term is formed by multiplying the preceding term by constant factor. This factor is called the common ratio. The sequence

3, 6, 12, 24, 48, 96

is an example of a geometric sequence. The 2nd term is equal to the 1st term times 2, the 3rd term is equal to the 2nd term times 2, the 4th term is equal to the 3rd term times 2, and so on. Notice that the sequence above is equal to

3, 3(2), 3(22), 3(23), 3(24), 3(25)

whose first term is 3 and common ratio 2.

In general, if the geometric sequence has first term a and common ratio r (not equal to 1), then the terms in the sequence are

a, ar, ar2, ar3, ar4

and so on, and the nth term is arn-1. This implies that if Sn is the sum of the geometric sequence, then

Sn =+ ar + ar2+ ar3+…+ arn-1.

Now, is there a shorter expression that is equivalent to the long addition above?

Yes, there is.

Notice that if we multiply the equation above by r, we have

rSn = ar + ar+ arar+… + arn.

If we subtract the rSfrom Sn, we have


Simplifying the equation, we have

$latex S_n (1-r) = a(1-r^n)$

which means that

$latex S_n = \displaystyle\frac{a(1-r^n)}{1-r}$.

Leave a Reply

Your email address will not be published. Required fields are marked *