Let a_n be the n-th term of an A.P.

Question

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

Accepted Answer

$$\sum_{r=1}^{100}a_{2r}=a_2+a_4+\cdots+a_{200}=\frac{100}{2}\big[2(a+d)+99(2d)\big]=\alpha$$
$$\sum_{r=1}^{100}a_{2r-1}=a_1+a_3+\cdots+a_{199}=\frac{100}{2}\big[2a+99(2d)\big]=\beta$$

Subtracting: $\dfrac{100}{2}(2d)=\alpha-\beta \Rightarrow 100d=\alpha-\beta \Rightarrow d=\dfrac{\alpha-\beta}{100}.$

Solutions · 1