Question

if a1,a2,a3,........,an are in AP, where ai>0 for all i, then find value of 1/root a1 +root a2 +i/root a2 +root a3 +.......+1/root an-1 + root an

Answer 1

This is a very important and challenging Question . It may come in boards

Answer 2

Answer: The answer is $\frac{n-1}{\sqrt{a_{n}} - \sqrt{a_{1}} }$

Step-by-step explanation:

Step 1:

let $a_{1} , a_{2}, ........ a_{n} be$ in AP

$d = a_2 - a_1 = a_3-a_2= ..............= a_n-a_{n-1} -------1\\ Now,\\\frac{1}{\sqrt{a_1} + \sqrt{a_2}}+\frac{1}{\sqrt{a_3} + \sqrt{a_4}}+.......+\frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}= \frac{1}{\sqrt{a_1} + \sqrt{a_2}}\frac{(\sqrt{a_1} - \sqrt{a_2})}{(\sqrt{a_1} - \sqrt{a_2})}+ ........+ \frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}\frac{(\sqrt{a_n} - \sqrt{a_{n-1}})}{(\sqrt{a_n} - \sqrt{a_{n-1}})}$

Step 2:

$\frac{1}{\sqrt{a_1} + \sqrt{a_2}}+\frac{1}{\sqrt{a_3} + \sqrt{a_4}}+.......+\frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}=\frac{\sqrt{a_n}-\sqrt{a_1}}{d} from1$

$\frac{1}{\sqrt{a_1} + \sqrt{a_2}}+\frac{1}{\sqrt{a_3} + \sqrt{a_4}}+.......+\frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}= \frac{{\sqrt{a_n}-\sqrt{a_1}}(\sqrt{a_n}+\sqrt{a_1})}{d(\sqrt{a_n}+\sqrt{a_1})}$

$\frac{1}{\sqrt{a_1} + \sqrt{a_2}}+\frac{1}{\sqrt{a_3} + \sqrt{a_4}}+.......+\frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}= \frac{a_n-a_1}{d(\sqrt{a_n}+\sqrt{a_1})}$

$\frac{1}{\sqrt{a_1} + \sqrt{a_2}}+\frac{1}{\sqrt{a_3} + \sqrt{a_4}}+.......+\frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}= \frac{a_1+(n-1)d - a_1}{d(\sqrt{a_n}+\sqrt{a_1})}\\\frac{1}{\sqrt{a_1} + \sqrt{a_2}}+\frac{1}{\sqrt{a_3} + \sqrt{a_4}}+.......+\frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}= \frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}$