Question

Show that the sum of ${(m+n)}^{th}$ and ${(m-n)}^{th}$ terms of an A.P. is equal to twice the $m^{th}$ term.

Answer 1

Let $a$ and $d$ be the first term and the common difference of A.P. respectively.
It is known that the $k^{th}$ term of an A.P. is given by
$a_k=a+(k-1)d\therefore a_{m+n}=a+(m+n-1)d{a}_{m-n}=a+(m-n-1)d{a}_m=a+(m-1)d\therefore a_{m+n}+a_{m-n}=a+(m+n-1)d+a+(m-n-1)d=2a+(m+n-1+m-n-1)d=2a+(2m-2)d=2a+2(m-1)d=2[a+(m-1\left)d\right]={2a}_m$
Hence proved.