Question
If $${a_1},{a_2},.....,{a_n}$$ are in H.P., then the expression $${a_1}{a_2} + {a_2}{a_3} + ..... + {a_{n - 1}}{a_n}$$ is equals to
A.
$$n\left( {{a_1} - {a_n}} \right)$$
B.
$$\left( {n - 1} \right)\left( {{a_1} - {a_n}} \right)$$
C.
$$n{a_1}{a_n}$$
D.
$$\left( {n - 1} \right){a_1}{a_n}$$
Answer :
$$\left( {n - 1} \right){a_1}{a_n}$$
Solution :
$$\eqalign{
& \frac{1}{{{a_2}}} - \frac{1}{{{a_1}}} = \frac{1}{{{a_3}}} - \frac{1}{{{a_2}}} = ...... = \frac{1}{{{a_n}}} - \frac{1}{{{a_{n - 1}}}} = d\,\,\,\left( {{\text{say}}} \right) \cr
& {\text{Then }}{a_1}{a_2} = \frac{{{a_1} - {a_2}}}{d},\,{a_2}{a_3} = \frac{{{a_2} - {a_3}}}{d},......,{a_{n - 1}}{a_n} = \frac{{{a_{n - 1}} - {a_n}}}{d} \cr
& \therefore \,\,{a_1}{a_2} + {a_2}{a_3} + ...... + {a_{n - 1}}{a_n} \cr
& = \frac{{{a_1} - {a_2}}}{d} + \frac{{{a_2} - {a_3}}}{d} + ...... + \frac{{{a_{n - 1}}{a_n}}}{d} \cr
& = \frac{1}{d}\left[ {{a_1} - {a_2} + {a_2} - {a_3} + ...... + {a_{n - 1}}{a_n}} \right] = \frac{{{a_1} - {a_n}}}{d} \cr
& {\text{Also, }}\frac{1}{{{a_n}}} = \frac{1}{{{a_1}}} + \left( {n - 1} \right)d \cr
& \Rightarrow \,\,\frac{{{a_1} - {a_n}}}{{{a_1}{a_n}}} = \left( {n - 1} \right)d \cr
& \Rightarrow \,\,\frac{{{a_1} - {a_n}}}{d} = \left( {n - 1} \right){a_1}{a_n} \cr
& {\text{Which is the required result}}{\text{.}} \cr} $$