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/rⁿ

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 - rⁿ)/(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 - rⁿ)/(1 - r)

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

Therefore,

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

= arⁿ

Now,

S_n = arⁿ (1 - rⁿ)/(1 - r)

Thus, the required ratio

S/S_n = [a (1 - rⁿ)/(1 - r)]/[arⁿ (1 - rⁿ)/(1 - r)]

= [a (1 - rⁿ)/(1 - r)] x [(1 - r)/arⁿ (1 - rⁿ)]

= 1/rⁿ

Hence Proved

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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/rⁿ

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/rⁿ