Let $\mathrm{S}=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots \cdots \mathrm{n}$ terms. Observe the following lists.
 

List-I	List-II
$\left(\mathrm{A}\right) \underset{\mathrm{n}\to \infty }{\mathrm{Lt}} \mathrm{S}=$	$\left(1\right) 0$
$\left(\mathrm{B}\right) \underset{\mathrm{n}\to 0}{\mathrm{Lt}} \mathrm{S}=$	$\left(2\right) \frac{3}{10}$
$\left(\mathrm{C}\right) \underset{\mathrm{n}\to 3}{\mathrm{Lt}} \mathrm{S}=$	$\left(3\right) \frac{1}{3}$
$\left(\mathrm{D}\right) \underset{\mathrm{n}\to 4}{\mathrm{Lt}} \mathrm{S}=$	$\left(4\right) 1$
	$\left(5\right) \frac{4}{13}$

The correct match is :  

$\begin{array}{rcl}\mathrm{S}& =& \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+......\\ {\mathrm{t}}_{\mathrm{n}}& =& \frac{1}{(3\mathrm{n}-2) (3\mathrm{n}+1)}\\ & =& \frac{1}{3} \left[\frac{1}{3\mathrm{n}-2}-\frac{1}{3\mathrm{n}+1}\right]\\ \mathrm{S}& =& \frac{1}{3} \left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.......+\frac{1}{3\mathrm{n}-2}-\frac{1}{3\mathrm{n}+2}\right]\\ \mathrm{S}& =& \frac{1}{3} \left(1-\frac{1}{3\mathrm{n}+1}\right)\end{array}$

1)  $\begin{array}{rcl}\underset{\mathrm{n}\to \infty }{\mathrm{lt}} \mathrm{S}& =& \frac{1}{3} \underset{\mathrm{n}\to \infty }{\mathrm{lt}} \left(1-\frac{1}{3\mathrm{n}+1}\right)\\ & =& \frac{1}{3} (1-0)=\frac{1}{3}\end{array}$  

2)  $\underset{\mathrm{n}\to 0}{\mathrm{lt}}=\mathrm{S}=\frac{1}{3} \underset{\mathrm{n}\to 0}{\mathrm{lt}} (1-1)=0$

3)  $\begin{array}{rcl}\underset{\mathrm{n}\to 3}{\mathrm{lt}} \mathrm{S}& =& \frac{1}{3} \underset{\mathrm{n}\to 3}{\mathrm{lt}} \left(1-\frac{1}{3\mathrm{n}+1}\right)\\ & =& \frac{1}{3} \left(1-\frac{1}{10}\right)=\frac{1}{3} \frac{9}{10}=\frac{3}{10}\end{array}$

D)  $\begin{array}{rcl}\underset{\mathrm{n}\to 4}{\mathrm{lt}} \mathrm{S}& =& \underset{\mathrm{n}\to 4}{\mathrm{lt}} \frac{1}{3}\left(1-\frac{1}{3\mathrm{n}+1}\right)\\ & =& \frac{1}{3}\left(1-\frac{1}{3}\right)\\ & =& \frac{1}{3}\left(\frac{12}{13}\right)\\ & =& \frac{4}{13}\end{array}$