If $a_{1}, a_{2} \ldots .$. are in A.P., then, $\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots .+\frac{1}{\sqrt{a_{n}}+\sqrt{a_{n+1}}}$ is equal to

What we know: $a_1, a_2, a_3, \ldots$ are in Arithmetic Progression (A.P.)

This means: $a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \ldots = d$ (common difference)

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The key technique is rationalization. For any general term $\frac{1}{\sqrt{a_r} + \sqrt{a_{r+1}}}$, multiply by the conjugate:

$\frac{1}{\sqrt{a_r} + \sqrt{a_{r+1}}} \times \frac{\sqrt{a_{r+1}} - \sqrt{a_r}}{\sqrt{a_{r+1}} - \sqrt{a_r}}$

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Using the identity $(x+y)(x-y) = x^2 - y^2$:

$(\sqrt{a_r} + \sqrt{a_{r+1}})(\sqrt{a_{r+1}} - \sqrt{a_r}) = a_{r+1} - a_r = d$

So each term becomes:

$\frac{1}{\sqrt{a_r} + \sqrt{a_{r+1}}} = \frac{\sqrt{a_{r+1}} - \sqrt{a_r}}{d}$

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Now we add all terms:

$\sum_{r=1}^{n} \frac{1}{\sqrt{a_r} + \sqrt{a_{r+1}}} = \frac{1}{d} \sum_{r=1}^{n} (\sqrt{a_{r+1}} - \sqrt{a_r})$

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This becomes a telescoping series:

$\frac{1}{d}[(\sqrt{a_2} - \sqrt{a_1}) + (\sqrt{a_3} - \sqrt{a_2}) + (\sqrt{a_4} - \sqrt{a_3}) + \ldots + (\sqrt{a_{n+1}} - \sqrt{a_n})]$

Most terms cancel out, leaving:

$\frac{1}{d}(\sqrt{a_{n+1}} - \sqrt{a_1})$

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For an A.P., we have $a_{n+1} - a_1 = nd$ and using the identity:

$a_{n+1} - a_1 = (\sqrt{a_{n+1}} - \sqrt{a_1})(\sqrt{a_{n+1}} + \sqrt{a_1})$

Therefore:

$nd = (\sqrt{a_{n+1}} - \sqrt{a_1})(\sqrt{a_{n+1}} + \sqrt{a_1})$

So:

$\frac{\sqrt{a_{n+1}} - \sqrt{a_1}}{d} = \frac{n}{\sqrt{a_1} + \sqrt{a_{n+1}}}$

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Final Answer:

$\frac{n}{\sqrt{a_1} + \sqrt{a_{n+1}}}$