Thread: Geometric sequence question.

The first term of a finite geometric series is 6 and the last term is 4374. The sum of all terms is 6558. Find the common ratio and the number of terms in the series.

So i get that T_{1= 6.}So, a=6. Sn= 6558. how to find r?

What is the formula for the sum of a geometric series?

If , , , ..., and , then, according to the standard formula for the sum of geometric progression, we get



from where

.

Not using any memorized formulas, just thinking about it:

Given that 6r^n = 4374, divide through by 6 to get r^n = 729. Note that 729 = 27^2, or 9^3, or 3^6; so the answer must be that either r=27 and n=2, r=9 and n=3, or r=3 and n=6. See which one results in a series whose sum of terms 6 + 6r + 6r^2 + ...6r^n = 6558.