Proof of a series involving Arithmetic Progression

Question

If $${a}_{1},{a}_{2},{a}_{3},.....{a}_{n-1},{a}_{n} $$ are in A.P., then show that $$ \frac{1}{{a}_{1}{a}_{n}} + \frac{1}{{a}_{2}{a}_{n-1}} +\frac{1}{{a}_{3}{a}_{n-2}} +.....+\frac{1}{{a}_{n}{a}_{1}} = \frac{2}{({a}_{1} + {a}_{n})} .\left(\frac{1}{{a}_{1}} + \frac{1}{{a}_{2}} + \frac{1}{{a}_{3}} +..... +\frac{1}{{a}_{n}}\right)$$

My Attempt: I tried to prove by assuming the common difference as d and considering the consecutive terms as $$ {a}_{2} = {a}_{1} + d , {a}_{3} = {a}_{1} + 2d,... $$& so on. But this approach didn't help me.

Any help will be appreciated.

Answer 1

Use $$a_1+a_n=a_2+a_{n-1}=...=a_{n}+a_1$$ and $$(x+y)\cdot\frac{1}{xy}=\frac{1}{x}+\frac{1}{y}.$$ Thus, $$(a_1+a_n)\left(\frac{1}{a_1a_n}+...+\frac{1}{a_na_1}\right)=...$$

Answer 2

Hint:  following up on OP's attempt:

I tried to prove by assuming the common difference as d and considering the consecutive terms

$$ \begin{align} \frac{1}{a_ka_{n-k+1}} &= \frac{1}{\big(a_1+(k-1)d\big)\big(a_1+(n-k)d\big)} \\ &=\frac{1}{2a_1+(n-1)d}\left(\frac{1}{a_1+(k-1)d}+\frac{1}{a_1+(n-k)d}\right) \\ &=\frac{1}{a_1+a_n}\left(\frac{1}{a_k}+\frac{1}{a_{n-k+1}}\right) \end{align} $$