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

Question

Let P= (3/17) + (33/172) + (333/173) + ....∞ then P equals (A) (3/17) (B) (7/17) (C) (7/17) (D) (51/112)

Tardigrade Admin · Accepted Answer

P = (3/17) +(33/172) +(333/173) + ....∞ ...(1) P × (1/17) = (3/172) +(33/173) +(333/174) + .....∞ ...(2) (1)-(2) gives P(1 -(1/17)) = (3/17) +(30/172) + (300/173) ....∞ = (3/17)(1 +(10/17) +(100/172) + ...∞) = (3/17) ((17/7)) (Using S∞ = (a/1 -r) for G.P.) or (16P/17) = (3/7) ∴ P= (51/112)

Q. Let $P = \frac{3}{17} + \frac{33}{1 7 ^{2}} + \frac{333}{1 7 ^{3}} + ....\infty$ then $P$ equals

$\frac{3}{17}$

$\frac{7}{17}$

$\frac{7}{17}$

$\frac{51}{112}$

Q. Let P=173​+17233​+173333​+....∞ then P equals

173​

177​

177​

11251​

P=173​+17233​+173333​+....∞...(1) P×171​=1723​+17333​+174333​+.....∞...(2) (1)−(2) gives P(1−171​)=173​+17230​+173300​....∞ =173​(1+1710​+172100​+...∞)=173​(717​) (Using S∞​=1−ra​ for G.P.) or 1716P​=73​∴P=11251​

Q. Let $P = \frac{3}{17} + \frac{33}{1 7 ^{2}} + \frac{333}{1 7 ^{3}} + ....\infty$ then $P$ equals

$\frac{3}{17}$

$\frac{7}{17}$

$\frac{7}{17}$

$\frac{51}{112}$