aligned & If a_1, a_2, a_n are in A.P. where a_i>0 for all i , show that \\ &1a_1+a_2+1a_2+a_3++1a_n-1+a_n &=n-1a_1+a_n aligned

Since a_1, a_2, a_n are in A.P. a_2-a_1=a_3-a_2==a_n-a_n-1=d Take 1a_1+a_2=a_2-a_1(a_2+a_1)(a_2-a_1) =a_2-a_1a_2-a_1=a_2-a_1d Similarly, 1a_2+a_3=a_3-a_2d And 1a_n-1+a_n=a_n-a_n-1d Adding, 1a_1+a_2+1a_2+a_3+... aligned & +1a_n-1+a_n=1d[a_n-a_1] \\ & =1d (a_n-a_1)(a_n+a_1)a_n+a_1 \\ & =1d(a_n-a_1a_n+a_1)=1d (n-1) da_n+a_1=n-1a_n+a_1 aligned

If a_1, a_2, a_n are in A.P. where a_i>0 for all i show that &\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}&=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}} : 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}

Explanation

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