Question

in an AP , prove that the sum of the terms equidistant from the beginning and the end is always same and sum is equal to sum of 1st and last term

Answer 1

Suppose a and d are the first term and the common difference of an A.P. respectively.
And suppose total number of terms is n and last term is l.
So number of terms equidistant from beginning and end is $\frac{n}{2}$.
And last term; l = $a+\left(n-1\right)d$ ...(i)
We know that sum of n number of terms = $\frac{n}{2}\left[2a+\left(n-1\right)d\right]$ = $\frac{n}{2}\left[a+a+\left(n-1\right)d\right]$ ...(ii)
So from (i) and (ii) we get;
Sum of terms = $\frac{n}{2}\left(a+l\right)$
Therefore the sum of the terms equidistant from the beginning and the end is always same and sum is equal to sum of 1st and last term.