Question

If $a_1, a_2, a_3, .... a_n$ are in A.P. where $a_i > 0$ for all i, then $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + … + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} =$

• A. $\frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$
• B. $\frac{n}{\sqrt{a_1} + \sqrt{a_n}}$
• C. $\frac{n+1}{\sqrt{a_1} + \sqrt{a_n}}$
• D. None of these

Answer 1

Answer: (A)

$\frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$

Answer 2

Let the given sum be $S$. The general term of the sum is $T_k = \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$. We rationalize the denominator of $T_k$: $T_k = \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \times \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{\sqrt{a_{k+1}} - \sqrt{a_k}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k}$ Since $a_1, a_2, ..., a_n$ are in Arithmetic Progression (A.P.), the common difference is $d = a_{k+1} - a_k$ for all $k$. So, $T_k = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$.

The sum $S$ is given by: $S = \sum_{k=1}^{n-1} T_k = \sum_{k=1}^{n-1} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} = \frac{1}{d} \sum_{k=1}^{n-1} (\sqrt{a_{k+1}} - \sqrt{a_k})$ This is a telescoping sum: $\sum_{k=1}^{n-1} (\sqrt{a_{k+1}} - \sqrt{a_k}) = (\sqrt{a_2} - \sqrt{a_1}) + (\sqrt{a_3} - \sqrt{a_2}) + ... + (\sqrt{a_n} - \sqrt{a_{n-1}})$ $= \sqrt{a_n} - \sqrt{a_1}$ Thus, $S = \frac{\sqrt{a_n} - \sqrt{a_1}}{d}$.

For an A.P., $a_n = a_1 + (n-1)d$. Therefore, the common difference $d$ can be expressed as: $d = \frac{a_n - a_1}{n-1}$ Substitute this expression for $d$ into the sum $S$: $S = \frac{\sqrt{a_n} - \sqrt{a_1}}{\frac{a_n - a_1}{n-1}} = \frac{(n-1)(\sqrt{a_n} - \sqrt{a_1})}{a_n - a_1}$ We can factor the denominator $a_n - a_1$ as a difference of squares: $a_n - a_1 = (\sqrt{a_n} - \sqrt{a_1})(\sqrt{a_n} + \sqrt{a_1})$. $S = \frac{(n-1)(\sqrt{a_n} - \sqrt{a_1})}{(\sqrt{a_n} - \sqrt{a_1})(\sqrt{a_n} + \sqrt{a_1})}$ Assuming $\sqrt{a_n} \neq \sqrt{a_1}$ (i.e. $d \neq 0$), we can cancel the term $(\sqrt{a_n} - \sqrt{a_1})$: $S = \frac{n-1}{\sqrt{a_n} + \sqrt{a_1}}$ The formula also holds when $d=0$. The condition $a_i > 0$ ensures that all square roots are real and positive, and the denominators are non-zero.