Question

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

A
317
B
717
C
37
D
51112

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)$$(2)-(1) \Rightarrow$$ $$P(1-\dfrac{1}{17}) = \dfrac{3}{17}+ \dfrac{30}{17^{2}}+\dfrac{300}{17^{3}}+.... \infty$$$$= \dfrac{3}{17} (1+\dfrac{10}{17}+\dfrac{100}{17^{2}}+... \infty$$$$= \dfrac{3}{17}(\dfrac{1}{1-\dfrac{10}{17}}) = \dfrac{3}{7}$$$$\Rightarrow \dfrac{16P}{17} = \dfrac{3}{7}$$ $$\therefore P= \dfrac{51}{112}$$Maths

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