• # question_answer If${{a}_{1}},{{a}_{2}},....{{a}_{n}}$are in HP, then the expression${{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+,....+{{a}_{n-1}}{{a}_{n}}$is equal to       AIEEE  Solved  Paper-2006 A) $(n-1)({{a}_{1}}-{{a}_{n}})$ B)        $n{{a}_{1}}{{a}_{n}}$   C)        $(n-1){{a}_{1}}{{a}_{n}}$            D)        $n({{a}_{1}}-{{a}_{n}})$

First, we will find d by using formula${{T}_{n}}=a+(n-1)d$and then determine the sum.              Since, ${{a}_{1}},{{a}_{2}},{{a}_{3}},....{{a}_{n}}$are in HP. $\therefore$ $\frac{1}{{{a}_{1}}},\frac{1}{{{a}_{2}}},\frac{1}{{{a}_{3}}},....\frac{1}{{{a}_{n}}}$are in AP. Let $d$be the common difference of AP. $\therefore$$\frac{1}{{{a}_{2}}}-\frac{1}{{{a}_{1}}}=d$ $\Rightarrow$${{a}_{1}}-{{a}_{2}}={{a}_{1}}{{a}_{2}}d$ Similarly,              ${{a}_{2}}-{{a}_{3}}={{a}_{2}}{{a}_{3}}d$ ????????? ????????? ${{a}_{n-1}}-{{a}_{n}}={{a}_{n-1}}{{a}_{n}}d$ On adding all the equations, we get ${{a}_{1}}-{{a}_{n}}=d\{{{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+......+{{a}_{n-1}}{{a}_{n}}\}$  ...(i) Also,   $\frac{1}{{{a}_{n}}}=\frac{1}{{{a}_{1}}}+(n-1)d$ $\Rightarrow$$d=\frac{{{a}_{1}}-{{a}_{n}}}{{{a}_{1}}{{a}_{n}}(n-1)}$ On putting this value of d in Eq. (i), we get ${{a}_{1}}-{{a}_{n}}=\frac{{{a}_{1}}-{{a}_{n}}}{{{a}_{1}}{{a}_{n}}(n-1)}\{{{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+.....+{{a}_{n-1}}{{a}_{n}}\}$ $\Rightarrow$${{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+....+{{a}_{n-1}}{{a}_{n}}={{a}_{1}}{{a}_{n}}(n-1)$