If $a_1,a_2,....,a_n$ are in A.P, then prove that:

Question

We have to prove that $\frac{1}{a_1a_n} +\frac{1}{a_2a_{n-1}}+....+\frac{1}{a_{n-1}a_2} +\frac{1}{a_aa_1} = \frac{2}{a_1+a_n}(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n})$

The middle term will depend on the fact that $n$ is even or odd. So, I can't understand how to proceed. If I add the common terms in $L.H.S$, then my final term will be the central term.

Answer 1

Note that $\frac 1{xy}=\frac 1{x+y}\left(\frac 1x+\frac 1y\right)$.

Note also that for an AP, $a_r+a_{n-r+1}=2m \; \scriptsize(r=1,2,\cdots, n)$ where $m$ is the "middle term". Hence $2m=a_1+a_n$.

$$\begin{align} \text{LHS}=\sum_{r=1}^n\frac 1{a_ra_{n-r+1}} &=\sum_{r=1}^m\frac 1{2(a_r+a_{n-r+1})}\left(\frac 1{a_r}+\frac 1{a_{n-r+1}}\right)\\ &=\frac 1{2m}\sum_{r=1}^m\left(\frac 1{a_r}+\frac 1{a_{n-r+1}}\right)\\ &=\frac 1{a_1+a_n}\left[\sum_{r=1}^m\frac 1{a_r}+\sum_{r'=1}^m\frac 1{a_r'}\right] && \scriptsize \text{(putting } r'=n-r+1)\\ &=\frac 1{a_1+a_n}\sum_{r=1}^m\left(\frac 1{a_r}+\frac 1{a_r}\right) && \scriptsize \text{(putting } r=r')\\ &=\frac 2{a_1+a_n}\cdot \sum_{r=1}^n\frac 1{a_r}\\ &=\text{RHS}\end{align}$$

Answer 2

Hint: Write $a_j = a+jd$ and see what happens.

Answer 3

Case 1: $n=2m-1$. Note that: $$\frac{1}{a_1a_{2m-1}}=\frac{1}{2a_m}\left(\frac{1}{a_{m}-(m-1)d}+\frac{1}{a_{m}+(m-1)d}\right).$$ Hence: $$\sum_{k=1}^n\frac 1{a_ka_{n-k+1}}=\sum_{k=1}^{2m-1}\frac 1{a_ka_{2m-k}}=\\ 2\cdot \frac{1}{2a_m}\sum_{k=1}^{m} \left(\frac{1}{a_{m}-(k-1)d}+\frac{1}{a_{m}+(k-1)d}\right)=\\ \frac{2}{a_1+a_{2m-1}}\sum_{k=1}^{2m-1} \frac{1}{a_{k}}=\frac{2}{a_1+a_n}\sum_{k=1}^n \frac{1}{a_k}.$$ Case 2: $n=2m$. You can do it similarly.