Question

A gp consists of an even number of terms (all terms being different). If the sum of all the terms is five times the sum of the terms occupying odd places, then the common ratio will be

Answer 1

Step-by-step explanation:

let total terms be 2n

let numbers be a, ar, ar^2....... ar^2n-2, ar^2n-1

Even G. P:- ar, ar^3,ar^5.......ar^2n-1

Odd G. P:- a, ar^2,ar^4.........ar^2n-2

sum of all terms=5*sum of terms at odd places

Sum of 2n = 5* sum of odd

=> a(r^2n-1) = 5* a(r^2n-1)

.............. ...............

r-1. r^2-1

=> {a2-b2=(a-b)(a+b)}

(r^2n-1) = 5* (r^2n-1)

............ ...............

r-1. (r-1)(r+1)

=> r^2n-1 = 5r^2n-5/ r+1

=> (r+1)(r^2n-1) = 5r^2n-5

=> (4-r)(r^2n-1) = 0

4-r=0

r=4