Question
In an A.P., show that $\text{a}_{\text{m}+\text{n}}+\text{a}_{\text{m}-\text{n}}=2\text{a}_\text{m}.$

Answer

It is given that the sequence $<\text{a}_\text{n}>$ is an A.P.
$\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}\ .....(1)$
Similarly from (1)
$\text{a}_{\text{m+n}}=\text{a}+(\text{m}+\text{n}-1)\text{d}\ .....(2)$
$\text{a}_{\text{m+n}}=\text{a}+(\text{m}-\text{n}-1)\text{d}\ .....(3)$
Adding (2) and (3)
$​​\text{a}_{​​\text{m+n}}+​​\text{a}​​_{\text{m}-​​\text{n}}=(​\text{a}+(​​\text{m+n}-1)​​\text{d})(​​\text{a}+(​​\text{m}-​​\text{n}-1)​​\text{d})$
$=2​​\text{a}+(​\text{m+n}-1+​​\text{m}-​​\text{n}-1)​​\text{d}$
$=​​\text{2a}+\text{2d}(​​\text{m}-1)$
$=2(​​\text{a}+(​​\text{m}-1)​​\text{d})$
$=​​\text{2a}_\text{m}$
Hence proved.

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