Question

If $S_1=a_2+a_4+a_6+...$ upto $100$ terms and $S_2=a_1+a_3+a_5+...$ upto $100$ terms of a certain A.P., then its common difference is

• A. $S_1-S_2$
• B. $S_2-S_1$
• C. $\frac{\left(S_1-S_2\right)}{2}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option

Step 1: Information required for the solution

As both $S_1\text{and}{S}_2$ are the sum of the same A.P.

We can see that $S_1$ is the sum of the even terms upto $100$ terms and $S_2$ is the sum of the odd terms upto $100$ terms.

Let us call the common difference as $d$ and it will be,

$d=a_2-a_1$

Step 2: Determination of common difference $\mathit{d}$,

For this, subtract $S_2$ from $S_1$, we get

$\begin{array}{rcl}{S}_1-S_2& =& a_2+a_4+a_6+...100\text{terms}-\left(a_1+a_3+a_5+...100\text{terms}\right)\\ & =& \left(a_2-a_1\right)+\left(a_4-a_3\right)+\left(a_6-a_5\right)+\dots 100\text{terms}\end{array}$

We know that all $\left(a_2-a_1\right),\left(a_4-a_3\right),\left(a_6-a_5\right),\text{and so on}$ are all $d$, then

$\begin{array}{rcl}{S}_1-S_2& =& d+d+d+\dots 100\text{terms}\\ & =& 100d\end{array}$

From this, we find that,

$d=\frac{S_1-S_2}{100}$

Hence, the correct option is (D).