Question

If $a_1,a_2,a_3,\dots ,a_n$ are in arithmetic progression, where $a_1>0$ for all i.
Prove that $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$

Answer 1

Since $a_1,a_2,a_3,\dots ,a_n$ are in AP.

$\therefore (a_2-a_1)=(a_3-a_2)=\dots =(a_n-a_{n-1})$ =d(say)

[$\because$ common difference of an AP are equal]

Now, LHS$=\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$

$=\frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\frac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+\dots +\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{a_n-a_{n-1}}$

[rationalise each term]

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

$=\frac{1}{d}[\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+\dots +\sqrt{a_n}-\sqrt{a_{n-1}}]$

$=\frac{1}{d}[\sqrt{a_n}-\sqrt{a_1}]=\frac{a_n-a_1}{d[\sqrt{a_n}+\sqrt{a_1}]}$

[$\because$ multiplying numerator and denominator by $\sqrt{a_n}+\sqrt{a_1}]$

$=\frac{[a_1+(n-1)d-a_1]}{d(\sqrt{a_n}+\sqrt{a_1})}[\because a_n=a_1+(n-1\left)d\right]$

$=\frac{(n-1)d}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$=RHS