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Question
If the sequence < an > is an A.P., show that am +n +am − n = 2am.
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Solution
Let the sequence < an > be an A.P. with the first term being A and the common difference being D.
To prove: am +n +am − n = 2am
LHS: am +n +am − n
\[= A + (m + n - 1)D + A + (m - n - 1)D { \because a_n = a + (n - 1)d}\]
\[ = A + mD + nD - D + A + mD - nD - D\]
\[ = 2A + 2mD - 2D . . . (i)\]
RHS: 2am
\[= 2[A + (m - 1)D]\]
\[ = 2A + 2mD - 2D . . . (ii)\]
From (i) and (ii), we get:
LHS = RHS
Hence, proved.
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