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Question

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


A
317
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B
717
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C
37
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D
51112
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Solution

The correct option is D $$\displaystyle \frac{51}{112}$$
$$ P = \dfrac{3}{17}+\dfrac{33}{17^{2}}+\dfrac{333}{17^{3}}+.... \infty $$_____(1)
$$ \dfrac{P}{17} = \dfrac{3}{17^{2}}+\dfrac{33}{17^{3}}+.... \infty $$_______(2)
$$ (2)-(1) \Rightarrow $$ 
$$ P(1-\dfrac{1}{17}) =  \dfrac{3}{17}+ \dfrac{30}{17^{2}}+\dfrac{300}{17^{3}}+.... \infty $$
$$ = \dfrac{3}{17} (1+\dfrac{10}{17}+\dfrac{100}{17^{2}}+... \infty $$
$$ = \dfrac{3}{17}(\dfrac{1}{1-\dfrac{10}{17}}) = \dfrac{3}{7} $$
$$ \Rightarrow \dfrac{16P}{17} = \dfrac{3}{7} $$ 
$$ \therefore P= \dfrac{51}{112} $$

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