The sum of first $n$ terms of $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots$ is

The sum of first $n$ terms of $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots$ is
  1. $\frac{3 n}{2(3 n+2)}$
  2. $\frac{3 n}{3 n+2}$
  3. $\frac{n}{2(3 n+2)}$
  4. $\frac{n}{3 n+2}$

Solution

$\begin{aligned} & \text { (c) Let, } S_n=\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots . . \text { upto } n \text { terms } \\ & =\frac{1}{3}\left[\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots . . \text { upto } n \text { terms }\right] \\ & =\frac{1}{3}\left[\frac{5-2}{2 \cdot 5}+\frac{8-5}{5 \cdot 8}+\frac{11-8}{8 \cdot 11}+\ldots . . \text { upto } n \text { terms }\right] \\ & =\frac{1}{3}\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)\right.\end{aligned}$ $\left.+\left(\frac{1}{3 n-1}-\frac{1}{3 n+2}\right)\right]$ $\begin{aligned} & =\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3 n+2}\right]=\frac{1}{3}\left[\frac{3 n+2-2}{2(3 n+2)}\right] \\ & \therefore \quad S_n=\frac{n}{2(3 n+2)}\end{aligned}$

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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