The ${\mathrm{n}}^{\mathrm{th}}$ term of the series $\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\dots \dots$is

$\begin{array}{rcl} & & \frac{1}{4·7}+\frac{1}{7·10}+\frac{1}{10·13}+.........\\ & & 4, 7, 10, ........ \mathrm{are} \mathrm{in} \mathrm{A}.\mathrm{P}\\ {\mathrm{t}}_{\mathrm{n}}& =& 4+(\mathrm{n}-3) 3=3\mathrm{n}+1\\ & & 7, 10, 13, ........ \mathrm{are} \mathrm{in} \mathrm{A}.\mathrm{P}\\ {\mathrm{t}}_{\mathrm{n}}& =& 7+(\mathrm{n}-1) 3=3\mathrm{n}+4\\ {\mathrm{T}}_{\mathrm{n}}& =& \frac{1}{(3\mathrm{n}+1) (3\mathrm{n}+4)}\\ & & \frac{1}{4·7}+\frac{1}{7·10}+\frac{1}{10·3}+........+\frac{1}{(3\mathrm{n}+1) (3\mathrm{n}+4)}\\ & =& \frac{1}{3} \left[\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.........+\frac{1}{3\mathrm{n}+1}-\frac{1}{3\mathrm{n}+4}\right]\\ & =& \frac{1}{3} \left[\frac{1}{4}-\frac{1}{3\mathrm{n}+4}\right]=\frac{3\mathrm{n}}{3\left(4\right) (3\mathrm{n}+4)}=\frac{\mathrm{n}}{4(3\mathrm{n}+4)}\end{array}$