Question

Let `a_1,a_2,a_3......a_n` be an A.P . Prove that: `1/(a_1*a_n) + 1/(a_2*a_(n-1)) + 1/(a_3*a_(n-2)) + ......... + 1/(a_n*a_1) = 2/(a_1 + a_n)[1/a_1 + 1/a_2 + 1/a_3 +....... + 1/a_n]`?

Answer 1

See below.

Explanation:

Assuming `a_0=0` for notational purposes, we have

`a_k = k delta` and then

`1/(a_k a_(n-k+1))=1/(k(n-k+1)delta^2)` and then

`sum_(k=1)^n1/(a_k a_(n-k+1)) = 1/delta^2 sum_(k=1)^n 1/(k(n-k+1))`

but `1/(k(n-k+1)) = 1/(n+1)(1/k+1/(n-k+1))` and then

and `sum_(k=1)^n (1/k+1/(n-k+1)) = 2 sum_(k=1)^n 1/k` and putting all together

`sum_(k=1)^n1/(a_k a_(n-k+1)) = 1/delta^2 2/(n+1) sum_(k=1)^n 1/k` but

`1/(a_1+a_n) = 1/delta 1/(n+1)` then

`sum_(k=1)^n1/(a_k a_(n-k+1)) = 2/(delta + n delta) sum_(k=1)^n 1/(k delta) = 2/(a_1+a_n)sum_(k=1)^n 1/a_k`