Let an be the nth term of an A.P. If ∑r=1100 a2r=α and ∑r=1100 a2r−1=β, then the common difference of the A.P. is:

Let $a_{n}$ be the $n t h$ term of an A.P. If $\sum_{r = 1}^{100} a_{2 r} = α$ and $\sum_{r = 1}^{100} a_{2 r - 1} = β$ , then the common difference of the A.P. is:

A
$α - β$
B
$\frac{1}{100} (α - β)$
C
$β - α$
D
$\frac{1}{200} (α - β)$

Solution:

Let $d$ the common difference of the A.P., then
$\begin{array}{l} α - β = \sum_{r = 1}^{100} (a_{2 r} - a_{2 r - 1}) = \sum_{r = 1}^{100} d = 100 d \\ \Rightarrow d = \frac{1}{100} (α - β) \end{array}$

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