If $a_1,a_2,\dots ,a_n$ are in A.P., then $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$ is equal to

Correct Option: 1

Let $d$ be the common difference of the Arithmetic Progression (A.P.). Therefore:
$a_2-a_1=a_3-a_2=\dots =a_n-a_{n-1}=d$

Rationalizing each term in the given series:
$\frac{1}{\sqrt{a_1}+\sqrt{a_2}}=\frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}=\frac{\sqrt{a_2}-\sqrt{a_1}}{d}$

Applying this rationalization to all terms, the sum $S$ becomes:
$S=\frac{1}{d}\left((\sqrt{a_2}-\sqrt{a_1})+(\sqrt{a_3}-\sqrt{a_2})+\dots +(\sqrt{a_n}-\sqrt{a_{n-1}})\right)$

Notice that all intermediate terms cancel out:
$S=\frac{1}{d}\left(\sqrt{a_n}-\sqrt{a_1}\right)$

We can multiply the numerator and the denominator by $\sqrt{a_n}+\sqrt{a_1}$ to rewrite this:
$S=\frac{1}{d}·\frac{a_n-a_1}{\sqrt{a_n}+\sqrt{a_1}}$

Since the terms are in A.P., we know that:
$a_n=a_1+(n-1)d$
$a_n-a_1=(n-1)d$

Substitute this back into our expression for $S$:
$S=\frac{1}{d}·\frac{(n-1)d}{\sqrt{a_n}+\sqrt{a_1}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$