If a1, a2, a3, …, an are in A.P.,

Solution:

Given $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are in A.P., where $a_{i}>0$ for all $i$ To prove that:

$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

$\Rightarrow \mathrm{LHS}=\frac{1}{\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{2}}}+\frac{1}{\sqrt{\mathrm{a}_{2}}+\sqrt{\mathrm{a}_{3}}}+\cdots+\frac{1}{\sqrt{\mathrm{a}_{\mathrm{n}-1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}}$

Multiplying the first term by $\frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{\sqrt{a_{1}}-\sqrt{a_{2}}}$, the second term by $\frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{\sqrt{a_{2}}-\sqrt{a_{3}}}$ and so on that is rationalizing each term

$\Rightarrow \mathrm{LHS}=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}} \times \frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{\sqrt{a_{1}}-\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}} \times \frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{\sqrt{a_{2}}-\sqrt{a_{3}}}+\cdots$ $+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \times \frac{\sqrt{a_{n-1}}-\sqrt{a_{n}}}{\sqrt{a_{n-1}}-\sqrt{a_{n}}}$

Now by using $(a+b)(a-b)=a^{2}-b^{2}$

$\Rightarrow$ LHS $=\frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{a_{1}-a_{2}}+\frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{a_{2}-a_{3}}+\cdots+\frac{\sqrt{a_{n-1}}-\sqrt{a_{n}}}{a_{n-1}-a_{n}}$

As $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are in AP let its common difference be $d$

$a_{2}-a_{1}=d, a_{3}-a_{2}=d \ldots a_{n}-a_{n-1}=d$

Hence multiplying by $-1$

$a_{1}-a_{2}=-d, a_{2}-a_{3}=-d \ldots a_{n}-a_{n-1}=-d$

Put these values in LHS

$\Rightarrow$ LHS $=\frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{-d}+\frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{-d}+\cdots+\frac{\sqrt{a_{n-1}}-\sqrt{a_{n}}}{a_{n-1}-a_{n}}$

$=-\frac{1}{d}\left(\sqrt{a_{1}}-\sqrt{a_{2}}+\sqrt{a_{2}}-\sqrt{a_{3}}+\cdots+\sqrt{a_{n-1}}-\sqrt{a_{n}}\right)$

$=-\frac{1}{d}\left(\sqrt{a_{1}}-\sqrt{a_{n}}\right)$

To the above equation multiply and divide by $\left(\sqrt{a_{1}}+\sqrt{a_{n}}\right)$

$\Rightarrow$ LHS $=-\frac{1}{d} \frac{\left(\sqrt{a_{1}}-\sqrt{a_{n}}\right)\left(\sqrt{a_{1}}+\sqrt{a_{n}}\right)}{\left(\sqrt{a_{1}}+\sqrt{a_{n}}\right)}$

Using $(a+b)(a-b)=a^{2}-b^{2}$

$\Rightarrow$ LHS $=-\frac{1}{d} \frac{a_{1}-a_{n}}{\left(\sqrt{a_{1}}+\sqrt{a_{n}}\right)}$

The $n^{\text {th }}$ term of $A P$ is given by $t_{n}=a+(n-1) d$

Where the $\mathrm{t}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}}$ is the last $\mathrm{n}^{\text {th }}$ term and $\mathrm{a}=\mathrm{a}_{1}$ is the first term Hence $a_{n}=a_{1}+(n-1) d$

$\Rightarrow a_{1}-a_{n}=-(n-1) d$

Substitute $a_{1}-a_{n}$ in LHS

$\Rightarrow \mathrm{LHS}=-\frac{1}{\mathrm{~d}} \frac{-(\mathrm{n}-1) \mathrm{d}}{\left(\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}\right)}$

$=\frac{(\mathrm{n}-1)}{\left(\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}\right)}$

$\Rightarrow \mathrm{LHS}=\mathrm{RHS}$

Hence proved