If $${a_r} > 0,r \in N$$   and $${a_1},{a_2},{a_3},.....,{a_{2n}}$$    are in A.P. then $$\frac{{{a_1} + {a_{2n}}}}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \frac{{{a_2} + {a_{2n - 1}}}}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + \frac{{{a_3} + {a_{2n - 2}}}}{{\sqrt {{a_3}} + \sqrt {{a_4}} }} + ..... + \frac{{{a_n} + {a_{n + 1}}}}{{\sqrt {{a_n}} + \sqrt {{a_{n + 1}}} }}$$             is equal to

Solution :
$${a_1} + {a_{2n}} = {a_2} + {a_{2n - 1}} = ..... = {a_n} + {a_{n + 1}} = k\left( {{\text{say}}} \right).$$
Expression $$ = k\left\{ {\frac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{{a_1} - {a_2}}} + ..... + \frac{{\sqrt {{a_n}} - \sqrt {{a_{n + 1}}} }}{{{a_n} - {a_{n + 1}}}}} \right\} = \frac{k}{{ - d}}\left( {\sqrt {{a_1}} - \sqrt {{a_{n + 1}}} } \right),$$
where $$d$$ = common difference
$$ = \frac{k}{{ - d}}\frac{{{a_1} - {a_{n + 1}}}}{{\sqrt {{a_1}} + \sqrt {{a_{n + 1}}} }} = \left( {{a_1} + {a_{2n}}} \right).\frac{{ - nd}}{{ - d\left( {\sqrt {{a_1}} + \sqrt {{a_{n + 1}}} } \right)}}.$$