If the sum to infinity of the series `3+(3+d)1/4+(3+2d)1/(4^2)+oo` is `(44)/9` , then find `ddot`

Explanation

`S=3+(3+d)1/4+(3+2d)1/4^(2)+..oo` (1)
`rArr1/4S=(3)1/4+(3+d)1/4^(2)+..oo` (2)
Subtracting (2) from (1), we have
`3/4S=3+(d)1/4+(d)1/4^(2)+..oo`
`=3+(d/4)/(1-1/4)`
`=3+d/3`
or `S=4+(4d)/9`
Given,
`4+(4d)/9=44/9`
or `(4d)/9=8/9`
or d=2