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 equal to
B
$$(n - 1)({a_1}\, - {a_n})$$
D
$$(n - 1)\,\,{a_1}{a_n}$$
CHECK ANSWER
Explanation $${1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d$$ (say)
Then $${a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......,\,{a_{n - 1}}{a_n} = {{{a_{n - 1}} - {a_n}} \over d}$$
$$\therefore$$ $${a_1}a{}_2 + {a_2}{a_3} + ......... + {a_{n - 1}}{a_n}$$
$$ = {{{a_1} - {a_2}} \over d} + {{{a_2} - {a_3}} \over d} + ......$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{{a_{n - 1}} - {a_n}} \over d}$$
$$ = {1 \over a}\left[ {{a_1}} \right. - {a_2} + {a_2} - {a_3} + .......$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + {a_{n - 1}} - an} \right] = {{{a_1} - {a_n}} \over d}$$
Also, $${1 \over {{a_n}}} = {1 \over {{a_1}}} + \left( {n - 1} \right)d$$
$$ \Rightarrow {{{a_1} - {a_n}} \over {{a_1}{a_n}}} = \left( {n - 1} \right)d$$
$$ \Rightarrow {{{a_1} - {a_n}} \over d} = \left( {n - 1} \right){a_1}{a_n}$$
Which is the required result.