# Show that t

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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