Consider an infinite geometric series with first term `a` and common ratio `r`. If its sum is 4 and the second term is 3/4, then (a)`a=4/7, r=3/7` (b). `a=2, r=3/8` (c). `a=3/2, r=1/2` (d). `a=3, r=1/4`

Options

A. `a=(4)/(7),r=(3)/(7)`
B. `a=2,r=(3)/(8)`
C. `a=(3)/(2),r=(1)/(2)`
D. `a=3,r=(1)/(4)`

Answer

D

Explanation

We have
`(a)/(1-r)=4andar=(3)/(4)`
`rArr" "a=4(1-r)andar=(3)/(4)`
`rArr" "4r(1-r)=(3)/(4)rArr r=(1)/(4),(3)/(4)`
When `r=(1)/(4),a=3` and when `r=(3)/(4),a=1`.
Hence, option (d) is correct.