If $\{a_{n}\}$ is an AP, then ${a_{1}}^{2} - {a_{2}}^{2} + {a_{3}}^{2} - {a_{4}}^{2} + . . . . . . . . . . + {a_{99}}^{2} - {a_{100}}^{2} =$

$\frac{50}{99} ({a_{1}}^{2} - {a_{100}}^{2})$

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$\frac{100}{99} ({a_{100}}^{2} - {a_{1}}^{2})$

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$\frac{50}{51} ({a_{1}}^{2} - {a_{100}}^{2})$

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none of these

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Solution

The correct option is A
$\frac{50}{99} ({a_{1}}^{2} - {a_{100}}^{2})$

Explanation for correct option:
Step-1: Simplify the given data.
Given, $a_{1}, a_{2}, a_{3}, a_{4} . . . . . . . . . . a_{n}$ are in A.P.
Let, common difference $d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3}$
$∴ {a_{1}}^{2} - {a_{2}}^{2} + {a_{3}}^{2} - {a_{4}}^{2} + . . . . . . . . . . + {a_{99}}^{2} - {a_{100}}^{2}$
$=$ $(a_{1} - a_{2}) (a_{1} + a_{2}) + (a_{3} - a_{4}) (a_{3} + a_{4}) + . . . . . . . . . . + (a_{99} - a_{100}) (a_{99} + a_{100})$
$=$ $(- d) (a_{1} + a_{2}) + (- d) (a_{3} + a_{4}) + . . . . . . . . . . + (- d) (a_{99} + a_{100})$
$=$ $(- d) (a_{1} + a_{2} + a_{3} + a_{4} + . . . . . . . . + a_{n})$
Step-2: Use formula $S_{n} = \frac{n}{2} (a + a_{n})$ to find the value:
$=$ $(- d) (\frac{100}{2} (a_{1} + a_{100}))$
Since, $a_{100} = a_{1} + 99 d \Rightarrow d = \frac{(a_{100} - a_{1})}{99}$
$=$ $(- \frac{(a_{100} - a_{1})}{99}) (\frac{100}{2} (a_{1} + a_{100}))$
$=$ $\frac{50}{99} [(a_{1} - a_{100}) (a_{1} + a_{100})]$
$=$ $\frac{50}{99} ({a_{1}}^{2} - {a_{100}}^{2})$
Hence, correct answer is option $(A)$

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If an is an AP, then a12-a22+a32-a42+..........+a992-a1002=

If $\{a_{n}\}$ is an AP, then ${a_{1}}^{2} - {a_{2}}^{2} + {a_{3}}^{2} - {a_{4}}^{2} + . . . . . . . . . . + {a_{99}}^{2} - {a_{100}}^{2} =$