Question

Solve the question given in attachment
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\red{\rm :\longmapsto\:a_1,a_2,a_3, - - - ,a_n \: are \: in \: AP \: and \: a_i > 0}$

Let assume that common difference of an AP is d

So, it implies

$\rm :\longmapsto\:a_2 - a_1 = d$

$\rm :\longmapsto\:a_3 - a_2 = d$

$\rm :\longmapsto\:a_4- a_3 = d$

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$\rm :\longmapsto\:a_n- a_{n - 1} = d$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{a_1} + \sqrt{a_2} }$

On rationalizing the denominator, we get

$\rm \:  =  \: \dfrac{1}{ \sqrt{a_1} + \sqrt{a_2} } \times \dfrac{ \sqrt{a_1} - \sqrt{a_2} }{ \sqrt{a_1} - \sqrt{a_2} }$

$\rm \:  =  \: \dfrac{ \sqrt{a_1} - \sqrt{a_2} }{a_1 - a_2}$

$\rm \:  =  \: \dfrac{ \sqrt{a_1} - \sqrt{a_2} }{ - d}$

$\rm \:  =  \: \dfrac{ \sqrt{a_2} - \sqrt{a_1} }{ d}$

Thus,

$\rm \: \dfrac{1}{ \sqrt{a_1} + \sqrt{a_2} } =  \: \dfrac{ \sqrt{a_2} - \sqrt{a_1} }{ d}$

Similarly,

$\rm \: \dfrac{1}{ \sqrt{a_2} + \sqrt{a_3} } =  \: \dfrac{ \sqrt{a_3} - \sqrt{a_2} }{ d}$

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$\rm \: \dfrac{1}{ \sqrt{a_{n - 1}} + \sqrt{a_n} } =  \: \dfrac{ \sqrt{a_n} - \sqrt{a_{n - 1}} }{ d}$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{a_1} + \sqrt{a_2} } + \dfrac{1}{ \sqrt{a_2} + \sqrt{a_3} } + - - - + \dfrac{1}{ \sqrt{a_{n - 1}} + \sqrt{a_n} }$

$\rm \:  =  \: \dfrac{ \sqrt{a_2} - \sqrt{a_1} }{ d} + \dfrac{ \sqrt{a_3} - \sqrt{a_2} }{ d} + - - + \dfrac{ \sqrt{a_n} - \sqrt{a_{n - 1}} }{ d}$

$\rm \:  =  \: \dfrac{1}{d}\bigg( \sqrt{a_2} - \sqrt{a_1} + \sqrt{a_3} - \sqrt{a_2} + - - + \sqrt{a_n} - \sqrt{a_{n - 1}} \bigg)$

$\rm \:  =  \: \dfrac{1}{d}\bigg( \sqrt{a_n} - \sqrt{a_1} \bigg)$

On rationalizing the numerator, we get

$\rm \:  =  \: \dfrac{ \sqrt{a_n} - \sqrt{a_1} }{d} \times \dfrac{ \sqrt{a_n} + \sqrt{a_1} }{ \sqrt{a_n} + \sqrt{a_1} }$

$\rm \:  =  \: \dfrac{ a_n - a_1 }{d[ \sqrt{a_n} + \sqrt{a_1}]}$

$\rm \:  =  \: \dfrac{ a + (n - 1)d - a }{d[ \sqrt{a_n} + \sqrt{a_1}]}$

$\rm \:  =  \: \dfrac{ (n - 1)d}{d[ \sqrt{a_n} + \sqrt{a_1}]}$

$\rm \:  =  \: \dfrac{ n - 1}{ \sqrt{a_n} + \sqrt{a_1}}$

Hence, Proved