Let \(a_n\) be the nth term of an AP. If \(\sum_{r=1}^{100}\) \(a_{2r}\) = \(\alpha\) and \(\sum_{r=1}^{100}\) \(a_{2r-1}\) = \(\beta\), then the common difference of the AP is

Solution :

Given, \(a_2 + a_4 + a_6 + …… + a_{200}\) = \(\alpha\)      ………(i)

and \(a_1 + a_3 + a_5 + ….. + a_{199}\) = \(\beta\)           ………(ii)

On subtracting equation (ii) from equation (i), we get

(\(a_2 – a_1\)) + (\(a_4 – a_3\)) + ……… + (\(a_{200} – a_{199}\)) = \(\alpha\) – \(\beta\)

\(\implies\) d + d + ……..+ 100 times =  \(\alpha\) – \(\beta\)

\(\implies\)  100d = \(\alpha\) – \(\beta\)

\(\therefore\)   d = \(\alpha – \beta\over 100\)

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