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The sum of 0.2 +0.22+0.222+ … to n ter...

The sum of ` 0.2 +0.22+0.222+ …` to n terms is equal to

A

`(2/9)- (2/81) (1-10^(-n))`

B

` n - (1/9 ) (1- 10^(-n))`

C

`(2/9) [ n-(1/9)(1-10^(-n))]`

D

` (2/9)`

Text Solution

Verified by Experts

The correct Answer is:
c
` = 0.2 +0.22+0.222 +…` up to n terms
` = 2 [ 0.1 +0.11 +0.111+…" up to n terms"` ]
` = 2/9 [ 0.9 + 0.99 + 0.999 + ..." up to n terms" ]`
` = 2/9 [ (1-0.1)+(1-0.1)^(2) +(1-0.1)^(3)+..." up to n terms " `
` =2/9 [ n-(0.1)+(0.1)^(2)+(0.1)^(3) +... "up to n terms "]`
` = 2/9 [ n -(0.1)(1-(0.1)^(n))/(1-0.1)] = 2/9 [n-(1/9) [1-10^(-n)]]`
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-SEQUENCES AND SERIES -EXERCISE 8
  1. The sum of 0.2 +0.22+0.222+ … to n terms is equal to

    Text Solution

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