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Question

Show that the sum of (m + n)th and (mn)th terms of an A.P. is equal to twice the mth term.

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Solution

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

amn = a + (mn –1) d

am = a + (m –1) d

am + n + amn = a + (m + n –1) d + a + (mn –1) d

= 2a + (m + n –1 + mn –1) d

= 2a + (2m – 2) d

= 2a + 2 (m – 1) d

=2 [a + (m – 1) d]

= 2am

Thus, the sum of (m + n)th and (mn)th terms of an A.P. is equal to twice the mth term.


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