Question

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

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

Answer 1

The correct option is A $\frac{n}{2n-1}(a_1^2-a_{2n}^2)$

Since, $a_1,a_2,a_3,.....a_{2n}$ form an AP.
Therefore,
$a_2-a_1=a_4-a_3=....a_{2n}-a_{2n-1}=d$
Here, $a_1^2-a_2^2+a_3^2-a_4^2+....+a_{2n-1}^2-a_{2n}^2$
$=(a_1-a_2)(a_1+a_2)+(a_3-a_4)(a_3+a_4)+....+(a_{2n-1}-a_{2n})\cdot (a_{2n-1}+a_{2n})$
$=-d(a_1+a_2+....+a_{2n})=-d\left(\frac{2n}{2}(a_1+a_{2n}\right)$
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
$=\frac{n(a_1-a_{2n})\cdot (a_1+a_{2n})}{2n-1}$
$=\frac{n}{2n-1}\cdot (a_1^2-a_{2n}^2)$.