Question

In an arithmetic progression $a_1,a_2,a_3,\dots$, sum $\left(S\right)={a_1}^2-{a_2}^2+{a_3}^2-{a_4}^2+\dots –{a_{2k}}^2$ is equal to:

• A. $\frac{k}{2k-1}\left({a_1}^2-{a_{2k}}^2\right)$
• B. $\frac{2k}{2k-1}\left({a_{2k}}^2-{a_1}^2\right)$
• C. $\frac{k}{2k-1}\left({a_{2k}}^2-{a_1}^2\right)$
• D. None of these.

Answer 1

The correct option is A

$\frac{k}{2k-1}\left({a_1}^2-{a_{2k}}^2\right)$

Explanation for the correct answer:

Step 1: Simplifying the given equation:

As $a_1,a_2,a_3,\dots a_{2k}$ are in AP, so

$a_2-a_1=a_3-a_2=a_4-a_3=a_{2k}-a_{2k-1}=d$

Now

$\begin{array}{rcl}S& =& \left(a_1+a_2\right)\left(a_1-a_2\right)+\left(a_3+a_4\right)\left(a_3-a_4\right)+\dots +\left(a_{2k-1}+a_{2k}\right)\left(a_{2k-1}–a_{2k}\right)\\ & =& -d\left(a_1+a_2+a_3+a_4+\dots +a_{2k-1}+a_{2k}\right)\end{array}$

Step 2: Apply the formula of $n^{th}$term of AP.

By using the formula of $n^{th}$term of AP, we get

$\begin{array}{rcl}{a}_1+\left(2k-1\right)d& =& a_{2k}\left[\because t_n=a+\left(n-1\right)d\right]\\ \Rightarrow \left(2k-1\right)d& =& a_{2k}-a_1\\ \Rightarrow d& =& \frac{a_{2k}-a_1}{2k-1}\end{array}$

Step 3: Find the value of $S$.

$\begin{array}{rcl}S& =& -\left(\frac{a_{2k}-a_1}{2k-1}\right)\left[\frac{2k}{2}\left\{2a_1+\left(2k-1)d\right)\right\}\right]\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{(}\mathbf{2}\mathbf{a}\mathbf{+}\left(n-1\right)\mathbf{d}\mathbf{)}\mathbf{]}\\ & =& \frac{a_1-a_{2k}}{2k-1}\left[\frac{2k}{2}\left(2a_1+a_{2k}-a_1\right)\right]\\ & =& \frac{k\left(a_1-a_{2k}\right)\left(a_1+a_{2k}\right)}{2k-1}\\ & =& \frac{k}{2k-1}\left({a_1}^2-{a_{2k}}^2\right)\end{array}$

Hence, option A is correct.