If {a_1},;{a_2},;{a_3}.......{a_n} are in A.P | vedclass

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Question

(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

Vedclass · Accepted Answer

$\frac{{n - 1}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }}$

Vedclass · Answer

(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}} }}$.

If ${a_1},\;{a_2},\;{a_3}.......{a_n}$ are in $A.P.$, where ${a_i} > 0$ for all $i$, then the value of $\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}} }} = $

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