Question

How many terms are to be added in the series $(8+6+4+\dots )$, to make the sum of $-52$ ?

• A. $3$
• B. $13$
• C. $12$
• D. $31$

Answer 1

Answer: (B)

$13$

Here $a_1=8$ and $d=-2$.

Hence

$S_n=\frac{n}{2}[2a_1+(n-1\left)d\right]$

$-52=\frac{n}{2}[16+(n-1\left)\right(-2\left)\right]$

$-52=n(8-(n-1\left)\right)$

$-52=n(9-n)$

$-52=9n-n^2$

$n^2-9n-52=0$

$(n-13)(n+4)=0$
$n=13$ and $n=-4$

Since number of terms cannot be a negative integer, hence number of terms is 13.