Question

If $\sum _{n=0}^{\infty }(-1{)}^n{x}^{n+1}=\frac{1}{10}$, then $x=$

• A. $\frac{1}{9}$
• B. $\frac{1}{10}$
• C. $\frac{1}{11}$
• D. $\frac{9}{10}$

Answer 1

The correct option is A $\frac{1}{9}$

$\sum _{n=0}^{\infty }(-1{)}^n{x}^{n+1}=\frac{1}{10}$

$\Rightarrow x-x^2+x^3-x^4+x^5-...=\frac{1}{10}$

The above expression is an infinite G.P. with common ratio of $(-x)$

Hence, $\frac{x}{1-(-x)}$ $=\frac{1}{10}$.

$\Rightarrow 10x=1+x$

Hence $x=\frac{1}{9}$