Question

Enter the finite geometric series from its given description, and then enter its sum.

A geometric series that starts with 4, ends with -12,500, and has a common ratio of -5.
The geometric series is _____
The sum of the geometric series is_____

Answer 1

`a_n= 4(-5)^((n-1))`

-10416

Explanation:

a = 4

r = -5

geometric series:
`a_n=ar^((n-1)) = 4(-5)^((n-1))`

Determine which term = -12500:

`a_n=-12500`

`4(-5)^((n-1))=-12500`

`(-5)^((n-1)) = -3125`

`(5)^((n-1)) = 3125`

`(n-1)ln(5) = ln(3125)`

`n-1 = ln(3125)/ln(5) = 5`

`n = 5 + 1 = 6`

`a_6=-12500`

Using geometric sum series formula:
`S_n=(a(1-r^n))/(1-r)`

Therefore,
`S_6=(4(1-(-5)^6))/(1-(-5)) =(-62496)/(6) =-10416`