Question

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...100terms$

• A. $\frac{25}{160}$
• B. $\frac{1}{6}$
• C. $\frac{25}{151}$
• D. $\frac{25}{152}$

Answer 1

The correct option is C

$\frac{25}{151}$

Explanation for the correct option:

Step 1. Find the general term of given series

Given series is $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...100terms$

Here $2,5,8\dots$are in A.P. with $a=2$ and $d=3$

General term of A.P. is$\Rightarrow T_n=2+(n-1)3=n-1$

and $5,8,11\dots$are in A.P. with $a=5$and $d=3$

General term of this A.P. is $\Rightarrow T_n=5+(n-1)3=3n+2$

Hence, general term or $n^{th}$of given series is $\frac{1}{(3n-1)(3n+2)}$

Step 2. Find the sum of required series by writing it as sum of difference of two terms

$S_n=\frac{1}{2\times 5}+\frac{1}{5\times 8}+\frac{1}{8\times 11}+\dots \frac{1}{(3n-1)(3n+2)}$

Multiply and divide $S_n$ by $3$

$\Rightarrow S_n=\frac{3}{3}\left[\frac{1}{2\times 5}+\frac{1}{5\times 8}+\frac{1}{8\times 11}+\dots \frac{1}{(3n-1)(3n+2)}\right]$

$\Rightarrow S_n=\frac{1}{3}\left[\frac{(5-2)}{2\times 5}+\frac{(8-5)}{5\times 8}+\frac{(11-8)}{8\times 11}+\dots \frac{(3n+2)-(3n-1)}{(3n-1)(3n+2)}\right]$

$\Rightarrow S_n=\frac{1}{3}\left[\frac{1}{2}–\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}+\dots \frac{1}{(3n-1)}–\frac{1}{(3n+2)}\right]$

$\Rightarrow S_n=\frac{1}{3}\left[\frac{1}{2}–\frac{1}{(3n+2)}\right]$ [All remaining terms cancelled out]

$\Rightarrow S_n=\frac{n}{2(3n+2)}$

Put $n=100$ in above equation

$\Rightarrow$Sum of $n$ terms of given series$=\frac{100}{2\times 302}=\frac{\mathbf{25}}{\mathbf{151}}$

Hence, Option ‘C’ is Correct.