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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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How many terms of the A.P. $-6,-\large\frac{11}{2}$$, -5........$ are needed to give the sum $-25$

$\begin{array}{1 1}20\;or\;5 \\ 25\;or \;4 \\ 21 \\ 24 \end{array} $

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1 Answer

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  • Sum of $n$ terms of an A.P.$=S_n=\large\frac{n}{2}$$[2a+(n-1)d]$
Given $A.P.$ is $-6,-\large\frac{11}{2}$$,-5........$
In this A.P, Common difference $d=-\large\frac{11}{2}$$-(-6)=\large\frac{1}{2}$
First term $a=-6$
Let the number of terms of the series be $n$
Also given that the sum of $n$ terms of the series$=-25$
$i.e.,S_n=-25$
We know that $S_n=\large\frac{n}{2}$$[2a+(n-1)d]$
$\Rightarrow\:-25=\large\frac{n}{2}$$[2\times (-6)+(n-1)\large\frac{1}{2}]$
$\Rightarrow\:-100=-25n+n^2$
$\Rightarrow\:n^2-25n+100=0$
$\Rightarrow\:(n-20)(n-5)=0$
$\Rightarrow\:n=5\:\:or\:\:20$
answered Feb 20, 2014 by rvidyagovindarajan_1
 

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