Sum to infinity
Level
pending
Geometric sequence in which T6=1. The ratio between the sum of first 4 terms to the sum of first and third terms is 3:2 . Find the sum to infinity
The answer is 64.
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1 solution
Let the first term be a & common ratio be r. Given T6=1 so ar^5=1. Now sum of first four terms=a+ar+ar^2+ar^3 & sum of first and third term =a+ar^2. According to question, (a+ar+ar^2+ar^3)/(a+ar^2)=3/2. Cross multiplying we get a 3rd degree equation in r , 2r^3-r^2+2r-1=0. Solving we get r=1/2, i, -i. So r=1/2. Putting this in ar^5=1, we get a=32. Finally sum upto infinity =a/(1-r) = 32/(1-1/2) =64.