If a_1, a_2, ... a_n are in A.P. where a_i>0 for all i , show that \ frac1sqrt(a_1)+sqrt(a_2)+frac1sqrt(a_2)+sqrt(a_3)+ * s+frac1sqrta_n-1+sqrt(a_n) =fracn-1sqrt(a_1)+sqrt(a_n) \

If a_1, a_2, a_n are in A.P. where a_i>0 for all i sh... | Sequence & Series (Mathematics)

Since

a_1, a_2, \ldots a_n

are in A.P.

a_2-a_1=a_3-a_2=\cdots=a_n-a_{n-1}=d

Take

\frac{1}{\sqrt{a_1}+\sqrt{a_2}}=\frac{\sqrt{a_2}-\sqrt{a_1}}{\left(\sqrt{a_2}+\sqrt{a_1}\right)\left(\sqrt{a_2}-\sqrt{a_1}\right)}

=\frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}=\frac{\sqrt{a_2}-\sqrt{a_1}}{d}

Similarly,

\frac{1}{\sqrt{a_2}+\sqrt{a_3}}=\frac{\sqrt{a_3}-\sqrt{a_2}}{d}

And

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

Adding,

\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots

...

\begin{aligned} & +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}=\frac{1}{d}\left[\sqrt{a_n}-\sqrt{a_1}\right] \\ & =\frac{1}{d} \frac{\left(\sqrt{a_n}-\sqrt{a_1}\right)\left(\sqrt{a_n}+\sqrt{a_1}\right)}{\sqrt{a_n}+\sqrt{a_1}} \\ & =\frac{1}{d}\left(\frac{a_n-a_1}{\sqrt{a_n}+\sqrt{a_1}}\right)=\frac{1}{d} \frac{(n-1) d}{\sqrt{a_n}+\sqrt{a_1}}=\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}} \end{aligned}