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જો $a_r > 0, r \in N$ અને $a_1$,$a_2$,$a_3$,..,$a_{2n}$ સમાંતર શ્રેણીમાં હોય,તો$\frac{{{a_1}\, + \,{a_{2n}}}}{{\sqrt {{a_1}} + \sqrt {{a_2}} }}\, + \,\frac{{{a_2}\, + \,{a_{2n - 1}}}}{{\sqrt {{a_2}} + \sqrt {{a_3}} }}\, + \,\frac{{{a_3}\, + \,{a_{2n - 2}}}}{{\sqrt {{a_3}} \, + \,\sqrt {{a_4}} }}\, + \,..\, + \,\frac{{{a_n}\, + \,{a_{n + 1}}}}{{\sqrt {{a_n}\,} \, + \,{a_{n + 1}}}}\, = \,.........$
$\frac{{n({a_1}\, - \,{a_{2n}})}}{{\sqrt {{a_1}} \, - \,\sqrt {{a_{n\, + \,1}}} }}$
$\frac{{n({a_1}\, + \,{a_{2n}})}}{{\sqrt {{a_1}} \, + \,\sqrt {{a_{n\, + \,1}}} }}$
$\frac{{n - 1}}{{\sqrt {{a_1}} \, + \,\sqrt {{a_{n + 1}}} }}$
આપેલ પૈકી એકપણ નહિ.
Solution
$a_1 + a_{2n} =a_2+ a_{2n-1}= … = a_n + a_{n+1 } = k$
સમીકરણ $ = \,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 {{{\text{a}}_{\text{1}}}} \, – \,\sqrt {{a_{n + 1}}} } \right),$
જ્યાં $d=$ સામાન્ય તફાવત
$ = \,\,\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)}}$
