Question

Let the sequence $a_1$ , $a_2$ , $a_3$ , $\dots$ , $a_{2n}$ form an $AP$ . Then, $a_1^2-a_2^2+a_3^2-\dots +a_{2n-1}^2-a_{2n}^2$ is

• A. $\frac{n}{2n-1}\left(a_1^2-a_{2n}^2\right)$
• B. $\frac{2n}{n-1}\left(a_{2n}^2-a_1^2\right)$
• C. $\frac{n}{n+1}\left(a_1^2+a_{2n}^2\right)$
• D. None of the above

Answer 1

Answer: (A)

Since, $a_1,a_2,a_3,\dots ,a_{2n}$ form an AP. Therefore, $a_2-a_1=a_4-a_3=\dots =a_{2n}-a_{2n-1}=d$
Here, $a_1^2-a_2^2+a_3^2-a_4^2+\dots +a_{2n-1}^2-a_{2n}^2$
$=\left(a_1-a_2\right)\left(a_1+a_2\right)+\left(a_3-a_4\right)\left(a_3+a_4\right)+\dots +\left(a_{2n-1}-a_{2n}\right)\cdot \left(a_{2n-1}+a_{2n}\right)$
$=-d\left(a_1+a_2+\dots +a_{2n}\right)=-d\left(\frac{2n}{2}\left(a_1+a_{2n}\right)\right)$
Also, we know $a_{2n}=a_1+(2n-1)d$
$\Rightarrow d=\frac{a_{2n}-a_1}{2n-1}$
$\Rightarrow -d=\frac{a_1-a_{2n}}{2n-1}$
$\therefore$ Therefore, the sum is
$=\frac{n\left(a_1-a_{2n}\right)\cdot \left(a_1+a_{2n}\right)}{2n-1}$
$=\frac{n}{2n-1}\cdot \left(a_1^2-a_{2n}^2\right)$