If ${\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,\dots \dots {\mathrm{a}}_{\mathrm{n}}$ be an A.P. of non-zero terms then $\frac{1}{{\mathrm{a}}_1{\mathrm{a}}_2}+\frac{1}{{\mathrm{a}}_2{\mathrm{a}}_3}+\dots \dots +\frac{1}{{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{a}}_{\mathrm{n}}}=$

 

$$\begin{gathered} \mathrm{d}={\mathrm{a}}_2-{\mathrm{a}}_1={\mathrm{a}}_3-{\mathrm{a}}_2={\mathrm{a}}_4-{\mathrm{a}}_3 ..... \\ \frac{1}{{\mathrm{a}}_1{\mathrm{a}}_2}=\frac{1}{\mathrm{a}(\mathrm{a}+\mathrm{d})}=\frac{1}{\mathrm{d}}\left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{a}+\mathrm{d}}\right] \\ \frac{1}{{\mathrm{a}}_2{\mathrm{a}}_3}=\frac{1}{(\mathrm{a}+\mathrm{d}) (\mathrm{a}+2\mathrm{d})}=\frac{1}{\mathrm{d}}\left[\frac{1}{\mathrm{a}+\mathrm{d}}-\frac{1}{\mathrm{a}+2\mathrm{d}}\right] \\ \frac{1}{{\mathrm{a}}_{\mathrm{n}-1} {\mathrm{a}}_{\mathrm{n}}}=\frac{1}{\left(\mathrm{a}+(\mathrm{n}-2)\mathrm{d}\right) \left(\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\right)}=\frac{1}{\mathrm{d}}\left[\frac{1}{\mathrm{a}+(\mathrm{n}-2)\mathrm{d}}-\frac{1}{\mathrm{a}+(\mathrm{n}-1)\mathrm{d}}\right] \\ \frac{1}{{\mathrm{a}}_1{\mathrm{a}}_2}+\frac{1}{{\mathrm{a}}_2{\mathrm{a}}_3}.....+\frac{1}{{\mathrm{a}}_{\mathrm{n}-1} {\mathrm{a}}_{\mathrm{n}}} \\ =\frac{1}{\mathrm{d}} \left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{a}+(\mathrm{n}-1)\mathrm{d}}\right]=\frac{1}{\mathrm{d}}\left[\frac{\mathrm{a}+(\mathrm{n}-1)\mathrm{d}-\mathrm{a}}{\mathrm{a}\left(\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\right)}\right] \\ =\frac{\mathrm{n}-1}{{\mathrm{a}}_1 {\mathrm{a}}_{\mathrm{n}}} \end{gathered}$$