Question

If a₁,a₂,..... are in A.P., then, \(\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + .... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}}\) is equal to

• A. \(\frac{n-1}{\sqrt{a_1}+\sqrt{a_{n-1}}}\)
• B. \(\frac{n}{\sqrt{a_1}+\sqrt{a_{n+1}}}\)
• C. \(\frac{n-1}{\sqrt{a_1}+\sqrt{a_{n}}}\)
• D. \(\frac{n}{\sqrt{a_1}-\sqrt{a_{n+1}}}\)

Answer 1

The Correct Option is B

Given that \(a_1, a_2, \ldots, a_n\) are terms of an arithmetic progression (A.P.). We need to evaluate the expression:

\(\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \ldots + \frac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}}\).

For an A.P., each term \(a_i\) can be expressed as \(a_i = a_1 + (i-1)d\) where \(d\) is the common difference. The expression given is a telescoping series. We shall use the identity for telescoping: \(\frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} = \sqrt{a_{i+1}}-\sqrt{a_i}\).

Applying this identity, the series becomes: 

\((\sqrt{a_2} - \sqrt{a_1}) + (\sqrt{a_3} - \sqrt{a_2}) + \ldots + (\sqrt{a_{n+1}} - \sqrt{a_n})\)

On simplifying, most terms cancel out due to the telescoping nature:

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

Notice the sum of telescoping terms reduces to this result over the given range.\(a_{n+1} = a_1 + nd\) represents the last term.

The sum of the series thus simplifies to:

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

Therefore, substituting this back into the context of the problem given the cancelation, this results in:

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

This choice matches the correct option provided:

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

This is the required sum of the series.