Question

if x1,x2,x3, ..... are in AP and xi>0 for all i , prove that 1/(rootx1 + rootx2) + 1/(rootx2 +rootx3) +...+ 1/((rootxn-1) + (rootn)) = (n-1)/(rootx1 + rootxn)

Answer 1

$\frac{1}{\sqrt{x_1} + \sqrt{x_2}} + \frac{1}{\sqrt{x_2}+ \sqrt{x_3}} +...+ \frac{1}{\sqrt{x_{n - 1}} + \sqrt{x_n}} \\ \\ = \frac{\sqrt{x_1} - \sqrt{x_2}}{(\sqrt{x_1} + \sqrt{x_2})(\sqrt{x_1} - \sqrt{x_2})} + \frac{\sqrt{x_2} - \sqrt{x_3}}{(\sqrt{x_2} + \sqrt{x_3})(\sqrt{x_2} - \sqrt{x_3})} +...+ \frac{\sqrt{x_{n - 1}} - \sqrt{x_n}}{(\sqrt{x_{n - 1}} + \sqrt{x_n})(\sqrt{x_{n - 1}} - \sqrt{x_n})}$

$\\ \\ = \frac{\sqrt{x_1} - \sqrt{x_2}}{(\sqrt{x_1})^2 - (\sqrt{x_2})^2}} + \frac{\sqrt{x_2} - \sqrt{x_3}}{(\sqrt{x_2})^2 - (\sqrt{x_3})^2}} +...+ \frac{\sqrt{x_{n - 1}} - \sqrt{x_n}}{(\sqrt{x_{n - 1}})^2 - (\sqrt{x_n})^2}} \\ \\ = \frac{\sqrt{x_1} - \sqrt{x_2}}{x_1 - x_2} + \frac{\sqrt{x_2} - \sqrt{x_3}}{x_2 - x_3} +...+ \frac{\sqrt{x_{n - 1}} - \sqrt{x_n}}{x_{n - 1} - x_n}$

$\\ \\ = \frac{\sqrt{x_1} - \sqrt{x_2}}{-d} + \frac{\sqrt{x_2} - \sqrt{x_3}}{-d} +...+ \frac{\sqrt{x_{n - 1}} - \sqrt{x_n}}{-d} \\ \\ = \frac{\sqrt{x_1} - \sqrt{x_2} + \sqrt{x_2} - \sqrt{x_3} + \sqrt{x_3} - \sqrt{x_4} +...+ \sqrt{x_{n - 1}} - \sqrt{x_n}}{-d} \\ \\ = \frac{\sqrt{x_1} - \sqrt{x_n}}{-d} \\ \\ = \frac{(\sqrt{x_1} - \sqrt{x_n})(\sqrt{x_1} + \sqrt{x_n})}{-d(\sqrt{x_1} + \sqrt{x_n})}$

$\\ \\ = \frac{(\sqrt{x_1})^2 - (\sqrt{x_n})^2}{-d(\sqrt{x_1} + \sqrt{x_n})} \\ \\ = \frac{x_1 - x_n}{-d(\sqrt{x_1} + \sqrt{x_n})} \\ \\ = \frac{(1 - n)d}{-d(\sqrt{x_1} + \sqrt{x_n})} \\ \\ = \frac{(1 - n)d \times (-1)}{-d(\sqrt{x_1} + \sqrt{x_n}) \times (-1)} \\ \\ = \frac{(1 - n)(-1)d}{(-d)(-1)(\sqrt{x_1} + \sqrt{x_n})} \\ \\ = \frac{(- 1 + n)d}{d(\sqrt{x_1} + \sqrt{x_n})} \\ \\ = \frac{- 1 + n}{\sqrt{x_1} + \sqrt{x_n}} \\ \\ = \frac{n - 1}{\sqrt{x_1} + \sqrt{x_n}}$

$\\ \\ \therefore\ \frac{1}{\sqrt{x_1} + \sqrt{x_2}} + \frac{1}{\sqrt{x_2} + \sqrt{x_3}} +...+ \frac{1}{\sqrt{x_{n - 1}} + \sqrt{x_n}} = \frac{n - 1}{\sqrt{x_1} + \sqrt{x_n}}$

$\\ \\ \\ Hope\ this\ may\ be\ helpful. \\ \\ Please\ mark\ my\ answer\ as\ the\ \bold{brainliest}\ if\ this\ may\ be\ helpful. \\ \\ Thank\ you.\ Have\ a\ nice\ day. \\ \\ \\ \#adithyasajeevan$