Find the sum of the following series up to n terms:
(i) 5 + 55 + 555 + ....      (ii) .6 +.66 +.666 + ....

Solution:

(i) 5 + 55 + 555 + ....

Let Sₙ = 5 + 55 + 555 + .... n terms

Sₙ = 5/9 [9 + 99 + 999 + .... n terms]

= 5/9 [(10 - 1) + (10² - 1) + (10³ - 1) .... n terms]

= 5/9 [(10 + 10² + 10³ + .... n terms) - n]

Now using the sum of n terms of GP,

= 5/9 [10 (10ⁿ - 1)/(10 - 1) - n]

= 5/9 [10 (10ⁿ - 1)/9 - n]

= 50/81 (10ⁿ - 1) - 5n/9

(ii) .6 +.66 +.666 + ....

Let Sₙ = 5.6 +.66 +.666 + .... n terms

Sₙ = 6 [.6 +.66 +.666 + .... n terms]

= 6/9 [0.9 + 0.99 + 0.999 + .... n terms]

= 6/9 [(1 - 1/10) + (1 - 1/10²) + (1 - 1/10³) + .... n terms]

= 2/3 [(1 + 1 + 1 + .... n terms) - 1/10 (1 + 1/10 + 1/10² + .... n terms]

Now using the sum of n terms of GP,

= 2/3 {n - 1/10 [(1 - (1/10)ⁿ/(1 - 1/10)]}

= 2n/3  - 2/30 x 10/9 (1 - 10^-n)

= 2n/3  - 2/27 (1 - 10^-n)

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NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 21

Find the sum of the following series up to n terms: (i) 5 + 55 + 555 + .... (ii) .6 +.66 +.666 + ....

Summary:

The sum of the first two series up to n terms are i) 50/81 (10ⁿ - 1) - 5n/9 and ii) 2n/3  - 2/27 (1 - 10^-n)