Question

Solve the sum of 8 terms in a geometric series?

Determine the sum of the first eight terms of the geometric series in which:

t1 = 42 and t9 = 10752

Answer 1

The sum of the first 8 terms is `10710`

Explanation:

The first step would be finding the common ratio.

If we take the geometric sequence with common ratio `2` and terms

`1, 2, 4, 8`

If given `t_1 = 1` and `t_4 = 8`, we can say that the common ratio is used `3` times, thus the common ratio is equivalent to `root(3)(8/1) = 2`, which is clear when we write out all `4` terms.

Applying this concept to our given problem, we see that

`r = root(8)(10752/42) = root(8)(256) = 2`

We know that the sum of a geometric series is given by

`S_n = (a(1 - r^n))/(1 - r)`

Applying this to our problem we get

`S_8 = (42(1 - 2^8))/(1 - 2)`

`S_8 = -42(1 - 2^8)`

`S_8 = 10710`

Hopefully this helps!