Solve the equation

Question:

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

  1. $\frac{\alpha-\beta}{200}$

  2. $\alpha-\beta$

  3. $\frac{\alpha-\beta}{100}$

  4. $\beta-\alpha$


Correct Option: , 3

Solution:

$\frac{100}{2}[\mathrm{a}+\mathrm{a}+198 \mathrm{~d}]=\beta \Rightarrow 50(2 \mathrm{a}+198 \mathrm{~d})=\beta \ldots \ldots(2)$

(1) $-(2) \quad \alpha-\beta=50(2 \mathrm{~d})=\mathrm{d}=\frac{\alpha-\beta}{100}$

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