Official solution
\[{1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d\] (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.