Show that thesum of (m+n) ^th and (m-n) ^th term of an A.P is equal to twice the m ^th term.

Let a be the first term and d be the common difference of an AP.

Then, 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 + ( m - 1 ) 2d

= 2 [ a + ( m - 1 ) d ]

= 2a_m

Hence, proved.