Question

The sum of how many terms of the series $6+12+18+24+.....$ is $1800$?

• A. $16$
• B. $24$
• C. $20$
• D. $18$
• E. $22$

Answer 1

The correct option is B $24$

This is an A.P in which $a=6,d=6$ and $S_n=1800$
Then, $\frac{n}{2}[2a+(n-1\left)d\right]=1800$
$\Rightarrow$ $\frac{n}{2}[2\times 6+(n-1)\times 6]=1800$
$\Rightarrow$ $3n(n+1)=1800$
$\Rightarrow$ $n(n+1)=600$
$\Rightarrow$ $n^2+n-600=0$
$\Rightarrow$$n^2+25n-24n-600=0$
$\Rightarrow$ $n(n+25)-24(n+25)=0$
$\Rightarrow$ $(n+5)(n-24)=0$
$\Rightarrow$ $n=24$
Number of terms $=24$