If an A.P. consists of $n$ terms with first term $a$ and $n$th term $l$ show that the sum of the $m$th term from the beginning and the $m$th term from the end is $(a + l)$

Given:

An A.P. consists of $n$ terms with first term $a$ and $n$th term $l$.

To do:

We have to show that the sum of the $m$th term from the beginning and the $m$th term from the end is $(a + l)$.

Solution:

First term $a_1=a$

nth term $a_n=l$

Let the common difference of the A.P. be $d$.

We know that,

$a_n=a+(n-1)d$

Therefore,

$m$th term from the beginning $a_m=a+(m-1)d$

$m$th term from the end $=l-(m-1)d$

Sum of the $m$th term from the beginning and the $m$th term from the end $=a+(m-1)d+l-(m-1)d$

$=a+l$

Hence proved.