UPSC NDA 2 2019 Mathematics Solved Paper

Let a be the first term and r be the c.r. of G.P. given the sum of odd terms of the G.P. = m
a+ar2+ar4+ar6+...+ar198=m
⇒a[1+r2+r4+r6+...+r198]=m
⇒a[1

{(r2)100−1}

r2−1

]=m
⇒a[

r200−1

r2−1

]=m....(i)
and the sum of even terms = n
[ar+ar3+ar5+...+ar199]=n
⇒ar[1+r2+r4+...+r198]=n
⇒ar[1

{(r2)100−1}

r2−1

]=n
⇒ar[

r200−1

r2−1

]=n.....(ii)
From, dividing (i) by (ii), we get

a[ r200−1 r2−1 ]

ar[ r200−1 r2−1 ]

=

m

n

⇒

1

r

=

m

n

⇒r=

n

m

Hence, common ratio =

n

m