If the sequence $⟨ a_{n} ⟩$ is in A.P., show that $a_{m + n +} + a_{m - n} = 2 a_{m}$

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Solution

$⟨ a_{n} ⟩$ is in A.P.
$a$ is the first term and $d$ is the common difference
$\begin{matrix} a_{m + n} = a + (m + n - 1) d a_{m - n} = a + (m - n - 1) d a_{m + n} + a_{m - n} = a + (m + n - 1) d + a + (m - n - 1) d = 2 a + d [(m + n - 1) + (m - n - 1)] = 2 a + d [2 m - 2] = 2 a + 2 d [m - 1] = 2 a_{m} \end{matrix}$

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