Correct option is (C) x
\(S_\infty=\frac{x}{2}+\frac{x}{4}+\frac{x}{8}+.......\)
\(=x\,(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.......)\)
\(=x\,(\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+.......)\)
\(=x\,\frac{\frac{1}{2}}{1-\frac{1}{2}}\) \((\because\) Sum of infinite terms in G.P. \(=\frac a{1-r}\) if r < 1)
\(=x\,\frac{\frac{1}{2}}{\frac{1}{2}}\) = x
\(\therefore\) \(\frac{x}{2} +\frac{x}{4} + \frac{x}{8} + ......=x\)
