Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Find the Sum of the Following Series: 0.5 + 0.55 + 0.555 + ... to N Terms. - Mathematics

Find the sum of the following series:

0.5 + 0.55 + 0.555 + ... to n terms.

#### Solution

We have,
0.5 + 0.55 + 0.555 + ... n terms

$S_n$ = 5 [0.1 + 0.11+0.111 + ... n terms]

$= \frac{5}{9}\left( 0 . 9 + 0 . 99 + 0 . 999 + . . . +\text { to n terms } \right)$

$= \frac{5}{9}\left\{ \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + . . . \text { n terms } \right\}$

$= \frac{5}{9}\left\{ \left( 1 - \frac{1}{10} \right) + \left( 1 - \frac{1}{100} \right) + \left( 1 - \frac{1}{1000} \right) + . . . \text { n terms } \right\}$

$= \frac{5}{9}\left\{ n - \left( \frac{1}{10} + \frac{1}{{10}^2} + \frac{1}{{10}^3} + . . . \text { n terms } \right) \right\}$

$= \frac{5}{9}\left\{ n - \frac{1}{10}\frac{\left( 1 - \left( \frac{1}{10} \right)^n \right)}{\left( 1 - \frac{1}{10} \right)} \right\}$

$= \frac{5}{9}\left\{ n - \frac{1}{9}\left( 1 - \frac{1}{{10}^n} \right) \right\}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Exercise 20.3 | Q 4.4 | Page 28