Question

For the geometric series `54+36+...+128/27`, how do you determine the sum?

Answer 1

`s_7 = 4118/27`

Explanation:

This is a geometric series with first term `54`, common difference `36/54 = 2/3` and `n = ?`.

Before finding the sum, we must find the number of terms, `n`.

`t_n = a xx r^(n - 1)`

`128/27 = 54 xx (2/3)^(n -1)`

`128/1458 = (2/3)^(n- 1)`

`64/729 = (2/3)^(n - 1)`

`2^6/3^6 = (2/3)^( n -1)`

`(2/3)^6 = (2/3)^(n - 1)`

We can eliminate the bases now since they're equivalent.

`6 = n - 1`

`n = 7`

The sum of n terms of a geometric series is given by `s_n = (a(1 - r^n))/(1 -r)`.

`s_7 = (54(1 - (2/3)^7))/(1- (2/3))`

`s_7 = (54(1 - 128/2187))/(1/3)`

`s_7 = (54(2059/2187))/(1/3)`

`s_7 = 3(54)(2059/2187)`

`s_7 = (333,558)/(2,187) = 4118/27 ~= 152.5`

Hopefully this helps!