If the sum to infinity of the series `1 + 4x + 7x^2 + 10x^3 +.............` is `35/16` then `x=` (A) `1/5` (B) `2/5` (C) `3/7` (D)`1/7`

Explanation

Let ` S_(infty)=1+4x+(7x)^(2) +(10x)^(3)+"...."" upto "infty ".....(i)"`
Multiplying both sides of Eq. (i) by x we get
` xS_(infty)=1+(4x)^(2)+(7x)^(3) +(10x)^(4)+"...."" upto "infty ".....(ii)"`
Subtracting Eq. (ii) from Eq. (i), we get
`(1- x)S_(infty)=1+3x+3x^(2)+3x^(3)+"...."" upto "infty`
`=1+3(x+x^(2)+x^(3)+"...."" upto "infty)=1+3((x)/(1-x))=((1+2x))/((1-x))`
`therefore S_(infty)=(1+2x)/(1-x)^(2)=(35)/(16) " " [therefore S_(infty)=(35)/(16)]`
`implies 16+32x=35-70x+35x^(2)`
`implies 35x^(2)-102x+19=0`
`implies (7x-19)(5x-1)=0`
`x ne (19)/(7) [ :. " for infinity series common ratio " -1ltxlt1]`
Hence, `x=(1)/(5)`