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# Sum of the series $\;S=\large\frac{4}{7}-\large\frac{5}{7^2}+\large\frac{4}{7^3}-\large\frac{5}{7^4}+.....\;\infty\;$ is :

$(a)\;\large\frac{4}{43}\qquad(b)\;\large\frac{3}{11}\qquad(c)\;\large\frac{23}{48}\qquad(d)\;\large\frac{23}{24}$

Can you answer this question?

Answer : (c) $\;\large\frac{23}{48}$
Explanation : $S=\large\frac{4}{7}-\large\frac{5}{7^2}+\large\frac{4}{7^3}-\large\frac{5}{7^4}+....\;\infty$
Explanation : $S=\large\frac{4}{7}-\large\frac{5}{7^2}+\large\frac{4}{7^3}-\large\frac{5}{7^4}+....\;\infty$
Explanation : $\large\frac{1}{7}S=\large\frac{4}{7^2}-\large\frac{5}{7^3}+\large\frac{4}{7^4}+....\;\infty$
$\large\frac{8}{7}S=\large\frac{4}{7}-\large\frac{1}{7^2}-\large\frac{1}{7^3}-\large\frac{1}{7^4}+...\;\infty$
$\large\frac{8}{7}S=\large\frac{4}{7}-\;[\large\frac{\large\frac{1}{7^2}}{1-\large\frac{1}{7}}]$
$\large\frac{8}{7}S=\large\frac{4}{7}-\large\frac{1}{4^2}$
$8S=4-\large\frac{1}{6}$
$S=\large\frac{23}{48}\;.$
answered Jan 23, 2014 by