The sum of the terms equidistant from the beginning and end in an A.P.

Let a1, a2, ............ an be an A.P with common difference d.

Then kth term is ak = a1 + (k – 1)d and kth term from last is the (n – k + 1)th term from beginning

i. e $a_{n-k+1}=a_{1}+(n-k+1-1) d$

$a_{n-k+1}=a_{1}+(n-k) d$

∴ kth term from beginning + kth term from the end is

$a_{1}+(k-1) d+a_{1}+(n-k) d$

$=2 a_{1}+d(k-1+n-k) d$

$=2 a_{1}+(n-1) d$

$=a_{1}+\left(a_{1}+(n-1) d\right)$

$=a_{1}+a_{n}$

∴ Sum of the terms equidistant from the beginning of an A.P and end of A.P is always same and is equal to sum of first term and nth term or last term of the A.P.