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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Sequences and Series

NDA 2018 · Mathematics · Sequences and Series

The sum of the series 3-1+(1)/(3)-(1)/(9)+ . is equal to

The sum of the series 3−1+13−19+…3-1+\frac{1}{3}-\frac{1}{9}+\ldots. is equal to

Options

  1. A209\frac{20}{9}
  2. B920\frac{9}{20}
  3. C94\frac{9}{4}
  4. D49\frac{4}{9}

Correct answer

C. 94\frac{9}{4}

Step-by-step solution

The series is 330−331+332−333+…∞\frac{3}{3^0}-\frac{3}{3^1}+\frac{3}{3^2}-\frac{3}{3^3}+\ldots \infty This is a G.P. with first term a=3a=3 and common ratio r=−13∴Sum⁡S=31−(−13)=3×33+1=94r=-\frac{1}{3}\therefore \operatorname{Sum} \mathrm{S}=\frac{3}{1-\left(-\frac{1}{3}\right)}=\frac{3 \times 3}{3+1}=\frac{9}{4}
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