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Question

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

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Solution

To prove 2am=am+n+amn

an=a+(n1)d

Similarly,

am+n=a+(m+n1)d

and amn=a+(mn1)d

am+n+amn

=a+(m+n1)d+a+(mn1)d

=2a+md+ndd+a+mdndd

=2a+2md2d

=2[a+(m1)d]

=2am

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