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Find the Sum of the Following Series: 0.6 + 0.66 + 0.666 + .... to N Terms - Mathematics

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Question

Find the sum of the following series:

0.6 + 0.66 + 0.666 + .... to n terms

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Solution

We have,
 0.6 + 0.66 +.666 + ... to n terms

\[S_n\] = 6 [0.1 + 0.11+ 0.111 + ... n terms]

\[= \frac{6}{9}\left( 0 . 9 + 0 . 99 + 0 . 999 + . . . \text { n terms } \right)\]

\[ = \frac{6}{9}\left\{ \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + . . .\text {  n terms } \right\}\]

\[ = \frac{6}{9}\left\{ \left( 1 - \frac{1}{10} \right) + \left( 1 - \frac{1}{100} \right) + \left( 1 - \frac{1}{1000} \right) + . . . \text { n terms } \right\} \]

\[ = \frac{6}{9}\left\{ n - \left( \frac{1}{10} + \frac{1}{{10}^2} + \frac{1}{{10}^3} + . . . \text { n terms } \right) \right\} \]

\[ = \frac{6}{9}\left\{ n - \frac{1}{10}\frac{\left( 1 - \left( \frac{1}{10} \right)^n \right)}{\left( 1 - \frac{1}{10} \right)} \right\}\]

\[ = \frac{6}{9}\left\{ n - \frac{1}{9}\left( 1 - \frac{1}{{10}^n} \right) \right\}\]

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Chapter 20: Geometric Progression - Exercise 20.3 [Page 28]

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RD Sharma Mathematics [English] Class 11
Chapter 20 Geometric Progression
Exercise 20.3 | Q 4.5 | Page 28

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