Question

How do you find `S_n` for the geometric series `a_3=-36`, `a_6=-972`, `n=10`?

Answer 1

The general form for a Geometric progression is:

`a_n = ar^(n-1)`

Given: `a_3 = -36` and `a_6 = -972`

Divide the latter by the former:

`a_6/a_3= (ar^(6-1))/(ar^(3-1)) = (-972)/-36 = 27`

This means that:

`r^3 = 27`

Solve for r:

`r = 3`

Use `a_3` and the value of r, to find the value of "a":

`a_3 = -36= a(3)^(3-1)`

`a(3)^2 = -36`

`a = -4`

We can use the formula for the Finite Sum of Geometric Series :

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

Substituting `a = -4, r = 3 and n = 10`

`S_n = -4(1-3^10)/(1-3)`

`S_n = 118096`