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 + ar¹ + ar² +....+ ar^{n - 1}

Given: Geometric sequence : a₁, a₂, a₃, a₄...... = -3, 18, -108, ...

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

To find r, 

calculate a₂/a₁ and a₃/a₁

⇒ 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 - rⁿ)/(1 - r), r ≠ 1

S₈ = (-3)[1 - (-6)⁸]/[1 - (-6)]

S₈ = (-3)[1 - (1679616)]/7

S₈ = (-3)(-1679615) / 7

S₈ = 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.