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.


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