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4. How many terms of the A.P. $-6 , -11/2 , -5...$ are needed to give the sum –25?

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Given : A.P. = $-6 , -11/2 , -5...$

$a=-6$

$d=\frac{-11}{2}+6=\frac{1}{2}$

Given : sum = -25

$S_n =\frac{n}{2}[2a+(n-1)d]$

$\Rightarrow \, \, -25=\frac{n}{2}[2(-6)+(n-1)\frac{1}{2}]$

$\Rightarrow \, \, \, \, -50= n[-12+(n-1)\frac{1}{2}]$

$\Rightarrow \, \, \, \, -50= -12n+ \frac{n^2}{2}-\frac{n}{2}$

$\Rightarrow \, \, \, \, -100= -24n+ n^2-n$

$\Rightarrow \, \, \, \, n^2-25n+100=0$

$\Rightarrow \, \, \, \, n^2-5n-20n+100=0$

$\Rightarrow \, \, \, \, n(n-5)-20(n-5)=0$

$\Rightarrow \, \, \, \, (n-5)(n-20)=0$

$\Rightarrow \, \, \, \, n=5\, \, or\, \, 20.$

Posted by

seema garhwal

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