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

Now the sum of first '2n' terms is

`S_(2n)=(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

`(S_(n))/(S_(2n)-S_(n))\\\\=((a(1-r^(n)))/(1-r))/((a(1-r^(2n)))/(1-r)-(a(1-r^(n)))/(1-r))\\\\=(1-r^(n))/(r^(n)-r^(2n))\\\\=(1-r^(n))/(r^(n)(1-r^(n)))\\\\=(1)/(r^(n))`