Let $${a_n}$$ be the $${n^{th}}$$ term of an A.P. If $$\sum\limits_{r = 1}^{100} {{a_{2r}} = \alpha } $$   and $$\sum\limits_{r = 1}^{100} {{a_{2r - 1}} = \beta }, $$   then the common difference of the A.P. is

Solution :
Let $$d$$ be the common difference of the A.P.
Then $${a_{2r}} = {a_{2r - 1}} + d.$$
$$\eqalign{ & \therefore \sum\limits_{r = 1}^{100} {{a_{2r}}} = \sum\limits_{r = 1}^{100} {\left( {{a_{2r - 1}} + d} \right)} = \sum\limits_{r = 1}^{100} {{a_{2r - 1}} + 100d} \cr & \Rightarrow \alpha = \beta + 100d \cr & \Rightarrow d = \frac{{\alpha - \beta }}{{100}} \cr} $$