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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Sequence and Series

Let $S=\frac{3}{19}+\frac{33}{(19)^2}+\frac{333}{(19)^3}+$......$\infty$. Find $S$.

$(a)\;\frac{19}{54}\qquad(b)\;\frac{3}{19}\qquad(c)\;\frac{27}{171}\qquad(d)\;None$

1 Answer

Answer : (a) $\frac{19}{54}$
Explanation : $S=\frac{3}{19}+\frac{33}{(19)^2}+\frac{333}{(19)^3}+\;....$
$\frac{1}{19}\;S=\frac{3}{(19)^2}+\frac{30}{(19)^2}+\frac{300}{(19)^3}+\;....$
$\frac{18}{19}\;S=\frac{\frac{3}{19}}{1-\frac{10}{19}}$
$S=\frac{1}{3}*\frac{19}{18}=\frac{19}{54}.$
answered Jan 21, 2014 by yamini.v
 

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