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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Sequences and Series

NDA 2017 · Mathematics · Sequences and Series

What is the sum of the series 0.3+0.33+0.333+ . n terms?

What is the sum of the series 0.3+0.33+0.333+….n0.3+0.33+0.333+\ldots. n terms?

Options

  1. A\frac{1}{3} \left[\mathrm{n}-\frac{1}{9} \llleft(1-\frac{1}{10^{\mathrm{n}}}\right)]
  2. B\frac{1}{3} \left[\mathrm{n}-\frac{2}{9} \llleft(1-\frac{1}{10^{\mathrm{n}}}\right)]
  3. C\frac{1}{3} \left[\mathrm{n}-\frac{1}{3} \llleft(1-\frac{1}{10^{\mathrm{n}}}\right)]
  4. D\frac{1}{3} \left[\mathrm{n}-\frac{1}{9} \llleft(1+\frac{1}{10^{\mathrm{n}}}\right)]

Correct answer

A. \frac{1}{3} \left[\mathrm{n}-\frac{1}{9} \llleft(1-\frac{1}{10^{\mathrm{n}}}\right)]

Step-by-step solution

0.3+0.33+0.333+…….n terms =310+33100+3331000+……=3(110+11100+1111000+…….)=39(910+99100+9991000+…….)=13[(1−110)+(1−1100)+(1−11000)+……..]=13[n−(110+1102+1103+……110n)]=13[n−110(1−110n)1−110]( Since, Sum of  n terms of G.P =a(1−rn)1−r]=13[n−10n−19(10n)]=13[n−19(1−110n)]\begin{aligned} & 0.3+0.33+0.333+\ldots \ldots . \mathrm{n} \text{ terms } \\ & =\frac{3}{10}+\frac{33}{100}+\frac{333}{1000}+\ldots \ldots \\ & =3\left(\frac{1}{10}+\frac{11}{100}+\frac{111}{1000}+\ldots \ldots .\right) \\ & =\frac{3}{9}\left(\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots \ldots .\right) \\ & =\frac{1}{3}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)+\ldots \ldots . .\right] \\ & =\frac{1}{3}\left[\mathrm{n}-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\ldots \ldots \frac{1}{10^{\mathrm{n}}}\right)\right] \\ & =\frac{1}{3}\left[n-\frac{\frac{1}{10}\left(1-\frac{1}{10^{\mathrm{n}}}\right)}{1-\frac{1}{10}}\right] \\ & \left(\text{ Since, Sum\text{ of } } n \text{ terms of G.P }=\frac{a\left(1-r^n\right)}{1-r}\right] \\ & =\frac{1}{3}\left[n-\frac{10^n-1}{9\left(10^n\right)}\right]=\frac{1}{3}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right] \end{aligned}
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