Question

How do you find `S_n` for the geometric series `a_2=-36`, a_5=972#, n=7?

Answer 1

`S_n=3^((n+1))(-1)^n-3`

Explanation:

The `n^(th)` term of a Geometric Series and its sum `S_n` up to n^(th)`term`a_n`, whose first term is`a_1`and common ratio is`r# is given by

`a_n=a_1xxr^((n-1))` and `S_n=a_1xx(r^n-1)/(r-1)`

as `a_2=a_1xxr=-36` and `a_5=a_1xxr^4=972`

Dividing latter by former, we get `r^3=-972/36=-27`

and hence `r=root(3)(-27)=-3`

and `a_1=36/-3=-12`

Hence, `S_n=-12xx((-3)^n-1)/((-3)-1)=-12xx((-3)^n-1)/(-4)` i.e.

`S_n=3((-3)^n-1)=3(3^n(-1)^n-1)=3^((n+1))(-1)^n-3`