Question

How many terms of the $A.P.-6,-\frac{11}{2},-5,....$ are needed to give the sum $-25$?

• A. $5$
• B. $55$
• C. $50$
• D. $25$

Answer 1

Answer: (A), (D)

$5$
$25$

Consider the given A.P. series $-6,-\frac{11}{2},-5,........$

Then

$a=-6$

$d=-\frac{11}{2}-(-6)$

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

$S_n=-25$

We know that,

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

$-25=\frac{n}{2}(2\times (-6)+(n-1)\times \frac{1}{2})$

$-25=\frac{n}{2}(-12+\frac{n}{2}-\frac{1}{2})$

$-50=n(-\frac{25}{2}+\frac{n}{2})$

$-50=n\left(\frac{-25+n}{2}\right)$

$-100=n(-25+n)$

$-100=-25n+n^2$

$n^2-25n+100=0$

$n^2-(20+5)n+100=0$

$n^2-20n-5n+100=0$

$n(n-20)-5(n-20)=0$

$(n-20)(n-5)=0$

If $n-20=0$ if $n-5=0$

Then,$n=20$ $n=5$

Hence, $n=5or20$ is the answer.