Sequences And Series Ques 2
- If the sum of the first ten terms of the series $\left(1 \frac{3}{5}\right)^2+\left(2 \frac{2}{5}\right)^2+\left(3 \frac{1}{5}\right)^2+4^2+\left(4 \frac{4}{5}\right)^2+\ldots$, is $\frac{16}{5} m$, then $m$ is equal to
(2016 Main)
(a) $102$
(b) $101$
(c) $100$
(d) $99$
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Answer:
Correct Answer: 2.(b)
Solution: (b) Let $S_{10}$ be the sum of first ten terms of the series. Then, we have
$ S_{10}=\left(1 \frac{3}{5}\right)^2+\left(2 \frac{2}{5}\right)^2+\left(3 \frac{1}{5}\right)^2+4^2+\left(4 \frac{4}{5}\right)^2 $ $+\ldots$ to $10$ terms
$ \begin{aligned} & =\left(\frac{8}{5}\right)^2+\left(\frac{12}{5}\right)^2+\left(\frac{16}{5}\right)^2+4^2+\left(\frac{24}{5}\right)^2+\ldots \text { to } 10 \text { terms } \\ & =\frac{1}{5^2}\left(8^2+12^2+16^2+20^2+24^2+\ldots \text { to } 10 \text { terms }\right) \\ & =\frac{4^2}{5^2}\left(2^2+3^2+4^2+5^2+\ldots \text { to } 10 \text { terms }\right) \\ & =\frac{4^2}{5^2}\left(2^2+3^2+4^2+5^2+\ldots+11^2\right) \\ & =\frac{16}{25}\left(\left(1^2+2^2+\ldots+11^2\right)-1^2\right) \\ & =\frac{16}{25}\left(\frac{11 \cdot(11+1)(2 \cdot 11+1)}{6}-1\right) \\ & =\frac{16}{25}(506-1)=\frac{16}{25} \times 505 \Rightarrow \frac{16}{5} m=\frac{16}{25} \times 505=101 \end{aligned} $
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