progressions

Problem 2:
sum series:
1/(1+√x) + 1/(1-x) + 1/(1-√x)+..........

(1):::$\textbf{If $\mathbf{a_{1},a_{2},a_{3},.........,{a_{n}}$ are in A.P, Then }$}$\\\\ $\mathbf{a_{1}+a_{n}=a_{2}+a_{n-1}=a_{3}+a_{n-2}=.......................=a_{n}+a_{1}=K}$\\\\ $\mathbf{\frac{1}{a_{1}.a_{n}}=\frac{1}{a_{1}+a_{n}}.\left(\frac{1}{a_{1}}+\frac{1}{a_{n}}\right)=\frac{1}{K}.\left(\frac{1}{a_{1}}+\frac{1}{a_{n}}\right)}$\\\\\\ $\mathbf{\frac{1}{a_{2}.a_{n-1}}=\frac{1}{a_{2}+a_{n-1}}.\left(\frac{1}{a_{2}}+\frac{1}{a_{n-1}}\right)=\frac{1}{K}.\left(\frac{1}{a_{2}}+\frac{1}{a_{n-1}}\right)}$\\\\\\ $\mathbf{\frac{1}{a_{3}.a_{n-2}}=\frac{1}{a_{3}+a_{n-2}}.\left(\frac{1}{a_{3}}+\frac{1}{a_{n-2}}\right)=\frac{1}{K}.\left(\frac{1}{a_{3}}+\frac{1}{a_{n-2}}\right)}$\\\\\\ .................................\\\\ ..................................\\\\ $\mathbf{\frac{1}{a_{n}.a_{1}}=\frac{1}{a_{n}+a_{1}}.\left(\frac{1}{a_{n}}+\frac{1}{a_{1}}\right)=\frac{1}{K}.\left(\frac{1}{a_{n}}+\frac{1}{a_{1}}\right)}$\\\\\\

\textbf{Now Add all These terms, we Get}$\\\\ $\mathbf{\frac{1}{a_{1}.a_{n}}+\frac{1}{a_{2}.a_{n-1}}+\frac{1}{a_{3}.a_{n-2}}+...............+\frac{1}{a_{n}.a_{1}}=\frac{2}{K}.\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+.........+\frac{1}{a_{n}}\right)}$\\\\\\ $\mathbf{\frac{1}{a_{1}.a_{n}}+\frac{1}{a_{2}.a_{n-1}}+\frac{1}{a_{3}.a_{n-2}}+...............+\frac{1}{a_{n}.a_{1}}=\frac{2}{a_{1}+a_{n}}.\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+.........+\frac{1}{a_{n}}\right)}$\\\\

i didnt get ur 2nd question !!