Show that in an arithmetical progression a1- a2- a3- S-a21-a22-a23-a24- a22k -k2k-1a21-a22k-

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Standard XII

Mathematics

nth Term of A.P

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Question

Show that in an arithmetical progression $a_{1}, a_{2}, a_{3}$ ,________,
$S = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} +$ ________ $a_{2 k}^{2}$
$= \frac{k}{2 k - 1} (a_{1}^{2} - a_{2 k}^{2})$ .

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Solution

$a_{1}^{2} - a_{2}^{2} = (a_{1} - a_{2}) (a_{1} + a_{2}) = - d (a_{1} + a_{2})$
Similarly for each of k brackets formed out of $2 k$ given terms.
$∴ S = - d S_{2 k} = - d \cdot \frac{2 k}{2} [a_{1} + a_{2 k}]$
$= - d k [\frac{a_{1}^{2} - a_{2 k}^{2}}{a_{1} - a_{2 k}}] = - d k \frac{(a_{1}^{2} - a_{2 k}^{2})}{a_{1} - {a_{1} + (2 k - 1) d}}$
$= \frac{k}{2 k - 1} (a_{1}^{2} - a_{2 k}^{2})$ .

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Show that in an arithmetical progression a1,a2,a3,________,S=a21−a22+a23−a24+________a22k=k2k−1(a21−a22k).

a21−a22=(a1−a2)(a1+a2)=−d(a1+a2)Similarly for each of k brackets formed out of 2k given terms.∴S=−dS2k=−d⋅2k2[a1+a2k]=−dk[a21−a22ka1−a2k]=−dk(a21−a22k)a1−{a1+(2k−1)d}=k2k−1(a21−a22k).

Show that in an arithmetical progression $a_{1}, a_{2}, a_{3}$ ,________,
$S = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} +$ ________ $a_{2 k}^{2}$
$= \frac{k}{2 k - 1} (a_{1}^{2} - a_{2 k}^{2})$ .