Question

(b) If a series follows geometric progression, show that the ratio of the sum of term to the sum from (n+1) term to (2n) term is 4 marks]

Answer 1

Proof with explanation:

We know that the sum of first 'n' terms of a Geometric progression is given by

$S_{n}=\frac{a(1-r^{n})}{1-r}$

where

a = first term of G.P

r is the common ratio

'n' is the number of terms

 Thus the sum of 'n' terms is

$S_{n}=\frac{a(1-r^{n})}{1-r}$

Now the sum of first '2n' terms is

$S_{2n}=\frac{a(1-r^{2n})}{1-r}$

Now the sum of terms from $(n+1)^{th}$ to $(2n)^{th}$ term is $S_{2n}-S_{n}$

Thus the ratio becomes

$\frac{S_{n}}{S_{2n}-S_{n}}\\\\=\frac{\frac{a(1-r^{n})}{1-r}}{\frac{a(1-r^{2n})}{1-r}-\frac{a(1-r^{n})}{1-r}}\\\\=\frac{1-r^{n}}{r^{n}-r^{2n}}\\\\=\frac{1-r^{n}}{r^{n}(1-r^{n})}\\\\=\frac{1}{r^{n}}$