Question

What is the sum of a 6–term geometric series if the first term is 20 and the last term is 336,140?

Answer 1

`392160`

Explanation:

Given:

`a_1 = 20`

`a_6 = 336140`

Then general form of a term of a geometric series is:

`a_n = a*r^(n-1)`

So `r^5 = (a r^5)/(a r^0) = a_6/a_1 = 336140/20 = 16807 = 7^5`

So assuming that the geometric series is of Real numbers, the only possible value of `r` is `7`.

Given any geometric series and positive integer `N`, we find:

`(r-1) sum_(n=1)^N a r^(n-1)`

`=r sum_(n=1)^N a r^(n-1) - sum_(n=1)^N a r^(n-1)`

`=ar^N + color(red)(cancel(color(black)(sum_(n=2)^(N-1) a r^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^(N-1) a r^(n-1)))) - a`

`=a(r^N-1)`

So dividing both ends by `(r-1)` we find:

`sum_(n=1)^N a r^(n-1) = (a(r^N-1))/(r-1)`

In our example, we have `a=20`, `r=7`, `N=6` and we get:

`sum_(n=1)^N a r^(n-1) = (a(r^N-1))/(r-1)`

`= (20(7^6-1))/(7-1) = (20(117649-1))/6 = 2352960/6 = 392160`