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.

Let $a$ and $d$ be the first term and the common difference of the $A.P.$ respectively.
It is known that the $k^{th}$ term of an $A.P.$ is given by $a_{k} = a + (k - 1)d$.
Therefore,$a_{m+n} = a + (m + n - 1)d$ and $a_{m-n} = a + (m - n - 1)d$.
Also,the $m^{th}$ term is $a_{m} = a + (m - 1)d$.
Now,the sum of the $(m+n)^{th}$ and $(m-n)^{th}$ terms is:
$a_{m+n} + a_{m-n} = [a + (m + n - 1)d] + [a + (m - n - 1)d]$
$= 2a + (m + n - 1 + m - n - 1)d$
$= 2a + (2m - 2)d$
$= 2a + 2(m - 1)d$
$= 2[a + (m - 1)d]$
$= 2a_{m}$
Thus,the sum of the $(m+n)^{th}$ and $(m-n)^{th}$ terms of an $A.P.$ is equal to twice the $m^{th}$ term.