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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If first and last term of an AP are $a$ & $l$ and sum of all terms is $s$, then common difference is

$(a)\;\frac{l^2+a^2}{s-(l-a)}\qquad(b)\;\frac{l^2+a^2}{2s-(l+a)}\qquad(c)\;\frac{l^2-a^2}{s-(l+a)}\qquad(d)\;\frac{l^2-a^2}{2s-(l+a)}$

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Answer : (d) $\;\frac{l^2-a^2}{2s-(l+a)}$
Explanation : $s=\frac{n}{2}\;(a+l)$
$n=\frac{2s}{(a+l)}$
$d=\frac{l-a}{n-1}=\frac{l-a}{\frac{2s}{(a+l)}-1}=\frac{l^2-a^2}{2s-(l+a)}.$
answered Jan 21, 2014 by yamini.v
 

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