# What is the sum of the geometric sequence -3, 18, -108, ... if there are 8 terms?

**Solution:**

Sum of the geometric sequence is :

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

Given: Geometric sequence : a_{1}, a_{2}, a_{3}, a_{4}...... = -3, 18, -108, ...

First term of the series a is -3, common ratio is r.

To find r,

calculate a_{2}/a_{1} and a_{3}/a_{1}

⇒ r = 18/(-3) = -108/18 = -6

⇒ r = -6

Since r < 1, sum of geometric sequence can be found by using the relation,

S_{n} = a(1 - r^{n})/(1 - r), r ≠ 1

S_{8} = (-3)[1 - (-6)^{8}]/[1 - (-6)]

S_{8} = (-3)[1 - (1679616)]/7

S_{8} = (-3)(-1679615) / 7

S_{8} = 719835

Therefore, the sum of geometric sequence if there are 8 terms is 719835.

## What is the sum of the geometric sequence -3, 18, -108, ... if there are 8 terms?

**Summary:**

The sum of the geometric sequence -3, 18, -108, ... if there are 8 terms is 719835.