An algebraic equality for arithmetic progressions Let $a_1, a_2, a_3,\ldots, a_n$ is an arithmetic progression . How can I prove that $\dfrac{1}{\sqrt{a_1}+\sqrt{a_2}}+\dfrac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots+\dfrac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}=\dfrac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$ .

HINT:
$$\frac1{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac1{\sqrt{a_k}+\sqrt{a_{k+1}}}\cdot\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{\sqrt{a_{k+1}}-\sqrt{a_k}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k}$$
Note that $a_{k+1}-a_k$ is just the constant difference of the arithmetic progression. The lefthand side telescopes, and you’ll have to perform a similar manipulation on the righthand side.
