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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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Knowledge Check

  • 0.2 + 0.22 + 0.222 + … upto 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)`
  • The sum of 0.2 + 0.22 + 0.222 +… upto n terms is equal

    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)`
  • If the sum to 2n terms of an Ap 2,5,8,11 , …. is equal to the sum to n terms of an AP 57,59,61,63, …. , then n is equal to

    A
    10
    B
    11
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    12
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