Question

Let the sequence $ a_{1} $ , $ a_{2} $ , $ a_{3} $ , $ \ldots $ , $ a_{2n} $ form an $ AP $ . Then, $ a^{2}_{1}-a^{2}_{2}+a^{2}_{3}-\ldots+a^{2}_{2n-1}-a^{2}_{2n} $ is

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

Answer 1

Answer: (A)

Since, $a_{1}, a_{2}, a_{3}, \ldots, a_{2 n}$ form an AP. Therefore, $a_{2}-a_{1}=a_{4}-a_{3}=\ldots=a_{2 n}-a_{2 n-1}=d$
Here, $a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-a_{4}^{2}+\ldots+a_{2 n-1}^{2}-a_{2 n}^{2}$
$=\left(a_{1}-a_{2}\right)\left(a_{1}+a_{2}\right)+\left(a_{3}-a_{4}\right)\left(a_{3}+a_{4}\right)+\ldots+\left(a_{2 n-1}-a_{2 n}\right) \cdot\left(a_{2 n-1}+a_{2 n}\right)$
$=-d\left(a_{1}+a_{2}+\ldots+a_{2 n}\right)=-d\left(\frac{2 n}{2}\left(a_{1}+a_{2 n}\right)\right)$
Also, we know $a_{2 n}=a_{1}+(2 n-1) d$
$\Rightarrow d=\frac{a_{2 n}-a_{1}}{2 n-1}$
$\Rightarrow -d=\frac{a_{1}-a_{2 n}}{2 n-1}$
$\therefore $ Therefore, the sum is
$=\frac{n\left(a_{1}-a_{2 n}\right) \cdot\left(a_{1}+a_{2 n}\right)}{2 n-1}$
$=\frac{n}{2 n-1} \cdot\left(a_{1}^{2}-a_{2 n}^{2}\right)$