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  4. Let P= (3/17) + (33/172) + (333/173) + ....∞ then P equals

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Q. Let $P= \frac{3}{17} + \frac{33}{17^{2}} + \frac{333}{17^{3}} + ....\infty$ then $P$ equals

Sequences and Series

Solution:

$P = \frac{3}{17} +\frac{33}{17^{2}} +\frac{333}{17^{3}} + ....\infty ...\left(1\right)$
$P \times \frac{1}{17} = \frac{3}{17^2} +\frac{33}{17^{3}} +\frac{333}{17^{4}} + .....\infty ...\left(2\right)$
$\left(1\right)-\left(2\right)$ gives
$P\left(1 -\frac{1}{17}\right) = \frac{3}{17} +\frac{30}{17^{2}} + \frac{300}{17^{3}} ....\infty$
$= \frac{3}{17}\left(1 +\frac{10}{17} +\frac{100}{17^{2}} + ...\infty\right) = \frac{3}{17} \left(\frac{17}{7}\right)$
(Using $S_{\infty} = \frac{a}{1 -r}$ for $G.P$.)
or $\frac{16P}{17} = \frac{3}{7} \therefore P= \frac{51}{112}$