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જો ${a_1},\;{a_2},\;{a_3}.......{a_n}$ એ સંમાતર શ્રેણીમંા હોય કે જયાંં ${a_i} > 0$,તો $\frac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \frac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + $ $........ + \frac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} = $ ___.
$\frac{{n - 1}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }}$
$\frac{{n + 1}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }}$
$\frac{{n - 1}}{{\sqrt {{a_1}} - \sqrt {{a_n}} }}$
$\frac{{n + 1}}{{\sqrt {{a_1}} - \sqrt {{a_n}} }}$
Solution
(a) As given ${a_2} – {a_1} = {a_3} – {a_2} = ……. = {a_n} – {a_{n – 1}} = d$
Where $d$ is the common difference of the given $A.P.$
Also ${a_n} = {a_1} + (n – 1)d$.
Then by rationalising each term,
$\frac{1}{{\sqrt {{a_2}} + \sqrt {{a_1}} }} + \frac{1}{{\sqrt {{a_3}} + \sqrt {{a_2}} }} + …. + \frac{1}{{\sqrt {{a_n}} + \sqrt {{a_{n – 1}}} }}$
$ = \frac{{\sqrt {{a_2}} – \sqrt {{a_1}} }}{{{a_2} – {a_1}}} + \frac{{\sqrt {{a_3}} – \sqrt {{a_2}} }}{{{a_3} – {a_2}}} + ….. + \frac{{\sqrt {{a_n}} – \sqrt {{a_{n – 1}}} }}{{{a_n} – {a_{n – 1}}}}$
$ = \frac{1}{d}\left\{ {\sqrt {{a_2}} – \sqrt {{a_1}} + \sqrt {{a_3}} – \sqrt {{a_2}} + …… + \sqrt {{a_n}} – \sqrt {{a_{n – 1}}} } \right\}$
$ = \frac{1}{d}\left\{ {\sqrt {{a_n}} – \sqrt {{a_1}} } \right\} = \frac{1}{d}\left( {\frac{{{a_n} – {a_1}}}{{\sqrt {{a_n}} + \sqrt {{a_1}} }}} \right)$
$ = \frac{1}{d}\left\{ {\frac{{(n – 1)d}}{{\sqrt {{a_n}} + \sqrt {{a_1}} }}} \right\} = \frac{{n – 1}}{{\sqrt {{a_n}} + \sqrt {{a_1}} }}$.
