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Question: The sum of 100 terms of the series <img src="https://cdn.pureessence.tech/canvas_65.png?top_left_x=1...

The sum of 100 terms of the series will be.

A

1−(110)1001 - \left( \frac { 1 } { 10 } \right) ^ { 100 }

B

1+(110)1001 + \left( \frac { 1 } { 10 } \right) ^ { 100 }

C

1−(110)1061 - \left( \frac { 1 } { 10 } \right) ^ { 106 }

D

1+(110)1001 + \left( \frac { 1 } { 10 } \right) ^ { 100 }

Answer

1−(110)1001 - \left( \frac { 1 } { 10 } \right) ^ { 100 }

Explanation

Solution

Series is a G.P.

With a=0.9=910a = 0.9 = \frac { 9 } { 10 } and r=110=0.1r = \frac { 1 } { 10 } = 0.1

∴\therefore S100=a(1−r1001−r)=910(1−1101001−110)=1−110100S _ { 100 } = a \left( \frac { 1 - r ^ { 100 } } { 1 - r } \right) = \frac { 9 } { 10 } \left( \frac { 1 - \frac { 1 } { 10 ^ { 100 } } } { 1 - \frac { 1 } { 10 } } \right) = 1 - \frac { 1 } { 10 ^ { 100 } }.