Let the sequence $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ form an A.P. Then
$a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + . . . . . + a_{2 n - 1}^{2} - a_{2 n}^{2}$ is equal to

$\frac{n}{2 n - 1} [a_{1}^{2} - a_{2 n}^{2}]$

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$\frac{n}{n - 1} [a_{0}^{2} - a_{2 n}^{2}]$

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$\frac{n}{n + 1} [a_{1}^{2} + a_{2 n}^{2}]$

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None of these

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Solution

The correct option is A
$\frac{n}{2 n - 1} [a_{1}^{2} - a_{2 n}^{2}]$

Suppose d is the common difference of the given A.P.
$d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} s = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + . . . . . . . a_{2 n - 1}^{2} . . . . . . . a_{2 n}^{2} = (a_{1} - a_{2}) (a_{1} + a_{2}) (a_{3} - a_{4}) (a_{3} + a_{4}) + . . . . . . . . (a_{2 n - 1} - a_{2 n}) (a_{2 n - 1} - a_{2 n}) = - d (a_{1} + a_{2} + a_{3} + . . . . . . + a_{2 n - 1} + a_{2 n}) = - d . \frac{2 n}{2} (a_{1} + a_{2} + . . . . . . . . . . + a_{2 n - 1} + a_{2 n}) = - d . \frac{2 n}{2} (a_{1} + a_{2 n}) = - n d (a_{1} + a_{2 n}) A l s o a_{2 n} = a_{1} + (2 n - 1) d \Rightarrow d = \frac{a_{2 n} - a_{1}}{2 n - 1} S = \frac{n}{2 n - 1} (a_{1} - a_{2 n}) (a_{1} + a_{2 n}) = \frac{n}{2 n - 1} (a_{1}^{2} - a_{2 n}^{2})$

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