Question

Let $a_n$ be the $n^{th}$ 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 an AP is

• A. $\frac{(\alpha -\beta )}{200}$
• B. $\frac{(\alpha -\beta )}{100}$
• C. $(\alpha -\beta )$
• D. $\beta -\alpha$

Answer 1

The correct option is B

$\frac{(\alpha -\beta )}{100}$

Explanation for the correct option:

Finding the common difference of the AP:

Given that,

$$\begin{gathered} \sum _{r=1}^{100}{a}_{2r}=\alpha \\ \Rightarrow a_2+a_4+a_6+\dots .+a_{200}=\alpha \dots \left(i\right) \end{gathered}$$

And $\sum _{r=1}^{100}{a}_{2r-1}=\beta$

$\Rightarrow a_1+a_3+a_5+\dots .+a_{199}=\beta \dots \left(ii\right)$

Subtracting equation $\left(i\right)$ from $\left(ii\right)$

$\Rightarrow (a_2-a_1)+(a_4-a_3)+(a_6-a_5)+\dots .+(a_{200}-a_{199})=\alpha -\beta$

Since the difference between two consecutive terms of AP is called common difference denoted by $d$ then

$$\begin{gathered} \Rightarrow d+d+d+\dots 100\text{times}=\alpha -\beta \\ \Rightarrow 100d=(\alpha -\beta ) \\ \Rightarrow d=\frac{(\alpha -\beta )}{100} \end{gathered}$$

Hence, the correct answer is option (B).