# 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}

**Solution:**

Let a be the first term and r be the common ratio of the G.P.

Sum of first n terms,

S = [a (1 - r^{n})/(1 - r)]

Since, there are n terms from (n + 1)^{th} to (2n)^{th} term.

Hence, sum of the n terms from (n + 1)^{th} to (2n)^{th} terms

S_{n} = a_{n + 1} (1 - r^{n})/(1 - r)

It is known that a_{n} = ar^{n }^{- 1}

Therefore,

a_{n + 1} = ar^{n + 1 - 1}

= ar^{n}

Now,

S_{n} = ar^{n} (1 - r^{n})/(1 - r)

Thus, the required ratio

S/S_{n} = [a (1 - r^{n})/(1 - r)]/[ar^{n} (1 - r^{n})/(1 - r)]

= [a (1 - r^{n})/(1 - r)] x [(1 - r)/ar^{n} (1 - r^{n})]

= 1/r^{n}

Hence Proved

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 24

## 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}

**Summary:**

Therefore, we have proved that the ratio of the sum of the first n terms of a G.P to the sum of terms from (n + 1)^{th} to (2n)^{th} term is 1/r^{n}