Question

Let the sequences $a_1,a_2,a_3,......,a_n,....$ from an AP. Then $a_1^2-a_2^2+a_3^2-a_4^2+...+a_{2n-1}^2-a_{2n}^2$ is equal to

• 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 these

Answer 1

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

Let the first term of the given A.P be $a$ and the common difference be $d$,

The corresponding terms will be $a+d,a+2d,a+3d,...,a+(2n-1)d$,

$\therefore {a_1}^2-{a_2}^2+{a_3}^2-{a_4}^2....+{a_{2n-1}}^2-{a_{2n}}^2⟹(a_1-a_2)(a_1+a_2)+(a_3-a_4)(a_3+a_4)+...+(a_{2n-1}-a_{2n})(a_{2n-1}+a_{2n})⟹(a_1+a_2)(-d)+(a_3+a_4)(-d)+...+(a_{2n-1}+a_{2n})(-d)⟹(-d)[a_1+a_2+a_3+...+a_{2n-1}+a_{2n}]⟹(-d)\left[\frac{2n}{2}(a_{2n}+a_1)\right]⟹(-d)\left[\frac{2n}{2}(a_{2n}+a_1)\right]\left[\frac{(a_{2n}-a_1)}{(a_{2n}-a_1)}\right]⟹\left[\frac{2n}{2}({a_{2n}}^2-{a_1}^2)\right]\frac{-d}{(2n-1)d}⟹\frac{n}{2n-1}({a_1}^2-{a_{2n}}^2)$