Explanation

(a)
Since, $a_{1},\ a_{2},\ a_{3},\ldots,a_{n}$ form an AP.
$\therefore\ a_{2} - a_{1} = a_{4} - a_{3} = ... = a_{2n} - a_{2n - 1} = d$
Let $S = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + ... + a_{2n - 1}^{2} - a_{2n}^{2}$
$= \left( a_{1} - a_{2} \right)\left( a_{1} + a_{2} \right) + \left( a_{3} - a_{4} \right)\left( a_{3} + a_{4} \right) + ... + \left( a_{2n - 1} - a_{2n} \right)(a_{2n - 1} + a_{2n})$
$= \ - d\left( a_{1} + a_{2} + ... + a_{2n} \right) = - d\left( \frac{2n}{2}\left( a_{1} + a_{2n} \right) \right)\$ …(i)
Also, we know $a_{2n} = a_{1} + (2n - 1)d$
$\Rightarrow \ d = \frac{a_{2n} - a}{2n - 1}\ \ \ \Rightarrow - d = \frac{a_{1} - a_{2n}}{2n - 1}$
On putting the value of $d$ in Eq, (i), we get
$S = \frac{n\left( a_{1} - a_{2n} \right)(a_{1} + a_{2n})}{2n - 1} = \frac{n}{2n - 1}(a_{1}^{2} - a_{2n}^{2})$