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Sequence

If the sequen...

Question

If the sequence < a_n > is an A.P., show that
a_m_+n +a_m_{− n} = 2a_m.

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Solution

Let the sequence < a_n > be an A.P. with the first term being A and the common difference being D.
To prove: a_m_+n +a_m_{− n} = 2a_m

LHS: a_m_+n +a_m_{− n

$= A + (m + n - 1) D + A + (m - n - 1) D {∵ a_{n} = a + (n - 1) d} = A + m D + n D - D + A + m D - n D - D = 2 A + 2 m D - 2 D . . . (i)$}

RHS: 2a_m

_{$= 2 [A + (m - 1) D] = 2 A + 2 m D - 2 D . . . (ii)$}

From (i) and (ii), we get:
LHS = RHS
Hence, proved.

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