# Show that the ratio of the sum of first n terms of a G.P to the sum of terms from (n+1)th to (2n)th term is 1/(r)n(1divided by r to the power n)..?please answer it as fast as possible...

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### 1 Answer

You need to remember how to evaluate the sum of n terms of geometric progression such that:

`S_n_1 = (a_1)*(r^n - 1)/(r-1)`

Notice that 1 represents the value of the first terms of geometric progression and r represents the common ratio.

You need to evaluate the sum of the n terms of the same geometric progression, having the first term `n+1` and the same common ratio r such that:

`S_n_2 = (a_(n+1))*(r^n - 1)/(r-1)`

You need to evaluate the following ratio such that:

`(S_n_1)/(S_n_2) = ((a_1)*(r^n - 1)/(r-1))/(a_(n+1)*(r^n - 1)/(r-1))`

Reducing like terms yields:

`(S_n_1)/(S_n_2) = (a_1)/(a_(n+1))`

You need to remember that you may write the `a_(n+1)` term using the first term and the common ratio such that:

`(a_(n+1)) = a_1*r^n`

`(S_n_1)/(S_n_2) = (a_1)/(a_1*r^n)`

Reducing like terms yields:

`(S_n_1)/(S_n_2) = 1/r^n`

**Hence, evaluating the ratio `(S_n_1)/(S_n_2)` yields `(S_n_1)/(S_n_2) = 1/r^n` .**