Let the sequence a_1, a_2, a_3, dots, a_{2n} | vedclass

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(A) Given that $a_1, a_2, a_3, \dots, a_{2n}$ form an $A.P.$ with common difference $d$.Therefore,$a_2 - a_1 = a_4 - a_3 = \dots = a_{2n} - a_{2n - 1}

Vedclass · Accepted Answer

$\frac{n}{2n - 1}(a_1^2 - a_{2n}^2)$

Vedclass · Answer

(A) Given that $a_1, a_2, a_3, \dots, a_{2n}$ form an $A.P.$ with common difference $d$.Therefore,$a_2 - a_1 = a_4 - a_3 = \dots = a_{2n} - a_{2n - 1} = d$.We need to evaluate the sum $S = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2n - 1}^2 - a_{2n}^2$.Using the difference of squares formula $x^2 - y^2 = (x - y)(x + y)$,we get:$S = (a_1 - a_2)(a_1 + a_2) + (a_3 - a_4)(a_3 + a_4) + \dots + (a_{2n - 1} - a_{2n})(a_{2n - 1} + a_{2n})$.Since $a_k - a_{k+1} = -d$ for odd $k$,we have:$S = -d(a_1 + a_2 + a_3 + a_4 + \dots + a_{2n - 1} + a_{2n})$.The sum of an $A.P.$ with $2n$ terms is $\frac{2n}{2}(a_1 + a_{2n}) = n(a_1 + a_{2n})$.Thus,$S = -d \cdot n(a_1 + a_{2n})$.From the formula for the $n$-th term,$a_{2n} = a_1 + (2n - 1)d$,so $d = \frac{a_{2n} - a_1}{2n - 1}$.Substituting $d$ into the expression for $S$:$S = -\left( \frac{a_{2n} - a_1}{2n - 1} \right) \cdot n(a_1 + a_{2n}) = \frac{n(a_1 - a_{2n})(a_1 + a_{2n})}{2n - 1} = \frac{n}{2n - 1}(a_1^2 - a_{2n}^2)$.

Let the sequence $a_1, a_2, a_3, \dots, a_{2n}$ form an $A.P.$ Then $a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2n - 1}^2 - a_{2n}^2 = $

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