Question

11. A geometric series consists of even number of terms. The sum of all terms is 3 times
the sum of odd terms. Find the common ratio.

Answer 1

Let a , ar , ar² , ar³ , ar⁴ , ar⁵
[ Note :- I assume just six terms in GP for simplicity . If want more terms to use here . You can use .]
terms which consist in even position are : ar , ar³ , ar⁵
terms which consist in odd position are : a , ar² , ar⁴
Now, sum of terms which consist even position = 3 × sum of terms which consist of position
[ar + ar³ + ar⁵] = 3 × [a + ar² + ar⁴ ]
ar[ 1 + r² + r⁴] = 3a[1 + r² + r⁴ ]
r = 3

Hence, common ratio = 3

Answer 2

HELLO DEAR,


let the terms be a, ar, ar², ar³, ar⁴,$ar^5$.....


now the term which consists in even is,
ar, ar³,$ar^5$.......

the term which consists in odd is,
a , ar² , ar⁴ ,..........


now,

The sum of all terms is 3*(the sum of odd terms)

(ar + ar³ + $ar^5$.......) = 3*(a + ar² + ar⁴.....)

ar(1 + r² + r⁴.....) = 3a(1 + r² + r⁴......)

r = 3



I HOPE ITS HELP YOU DEAR,
THANKS