Question

If the sum to infinity of the series $1+4x+7x^2+10x^3+.\dots \dots .$ is $\frac{35}{16}$ , where $\left|x\right|<1$ , then $x$ is equal to

• A. $\frac{19}{7}$
• B. $\frac{1}{5}$
• C. $\frac{1}{4}$
• D. $\frac{4}{7}$

Answer 1

Answer: (B)

$S=1+4x+7x^2+10x^3+.\dots \dots .$
$xS=x+4x^2+7x^3+.\dots \dots ..$
On subtracting, we get,
$S\left(1-x\right)=1+3x+3x^2+3x^3+.\dots \dots ..$
$=1+3x\left(\frac{1}{1-x}\right),\left|x\right|<1$
$\left(1-x\right)S=\frac{1-x+3x}{1-x}=\frac{1+2x}{1-x}$
$S=\frac{1+2x}{{\left(1-x\right)}^2}=\frac{35}{16}$ (given)
$16+32x=35+35x^2-70x$
$\Rightarrow 35x^2-102x+19=0$
$\Rightarrow \left(5x-1\right)\left(7x-19\right)=0$
$x=\frac{1}{5},\frac{19}{7}$
But, $\left|x\right|<1\Rightarrow x=\frac{1}{5}$