Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Let a and d be the first term and the common difference of the A.P. respectively.
It is known that the kth term of an A.P. is given by
ak = a + (k –1) d
∴ am + n = a + (m + n –1) d
am – n = a + (m – n –1) d
am = a + (m –1) d
Now,
L.H.S = am + n + am – 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]
= 2am
= R.H.S
Thus, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.