Let the sequence a1-a2-a3- form an A-P- then the value of a12-a22-a32-a42-a2n-12-a2n2 is equal to

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Sum of n Terms

Let the seque...

Question

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

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

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

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\frac{n}{n + 1} (a_{1}^{2} + a_{2 n}^{2})

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Solution

The correct option is C $\frac{n}{n + 1} (a_{1}^{2} + a_{2 n}^{2})$
Let $d$ be the common difference of A.P
Now $a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + . . . + a_{2 n - 1}^{2} - a_{2 n}^{2}$

$= (a_{1} - a_{2}) (a_{1} + a_{2}) + (a_{3} - a_{4}) (a_{3} + a_{4}) + . . . + (a_{2 n - 1} - a_{2 n}) (a_{2 n - 1} + a_{2 n})$

We know that, $a_{2} - a_{1} = d ∴ a_{1} - a_{2} = - d$
$= (- d) [a_{1} + a_{2} . . . + a_{2 n}]$
$= (- d) (\frac{2 n}{2}) [a_{1} + a_{2 n}]$
$= - n d (a_{1} + a_{2 n})$
$= - n [\frac{a_{2 n} - a_{1}}{2 n - 1}] (a_{1} + a_{2 n})$ $⎡ ⎢ ⎣ \begin{matrix} a s a_{2 n} = a_{1} + (2 n - 1) d ∴ d = \frac{a_{2 n} - a_{1}}{2 n - 1} \end{matrix} ⎤ ⎥ ⎦$
$= \frac{- n}{2 n - 1} (a_{2 n}^{2} - a_{1}^{2})$

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

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Similar questions

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

Q. Let

a_{n}

be the

n^{t h}

term of an

A P

.Show that

{a_{1}}^{2} - {a_{2}}^{2} + {a_{3}}^{2} - {a_{4}}^{2} + . . . . . . . . . . + {a^{2}}_{2 n - 1} - {a^{2}}_{2 n} = \frac{n}{2 n - 1} ({a_{1}}^{2} - {a_{2 n}}^{2})

Q. Let the sequences

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from an AP. 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

Q. If

a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + . . . . . + a_{2 n}^{2} = k a_{n}

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Let the sequence a1,a2,a3... form an A.P. then the value of a21−a22+a23−a24+⋯+a22n−1−a22n is equal to

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