Question

If the sum to infinity of the series $1+4x+7x^{2}+10x^{3}+.\ldots \ldots .$ 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)

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