Question

if the third term of gp is 2/9 and sum of infinite terms is 3/2 then find the series

Answer 1

let first term of gp be a₁ and r be the common ratio then
3rd term=a₁r²
a₁r²=2/9
or a₁=2/9r²

sum of infinite series of gp when -1<r<1

Sum=a₁/(1-r)
3/2=2/9r²(1-r)
or r²(1-r)=4/27
r²(1-r)=(2/3)² x 1/3

r²(1-r)=(2/3)² x (1-2/3)
hence r =2/3
then a₁=2/9 /(2/3)²=2*9/(9*4)
a₁=1/2
series is 
1/2, 1/3,2/9,4/27,8/81,.......