Question

Find the least value of n for which the series $3+6+9+....\dots$ upto $n$ terms exceeds $1000$.

• A. $20$
• B. $24$
• C. $25$
• D. $26$

Answer 1

The correct option is D $26$

Let $S_n=3+6+9+....\dots$(n terms)

$\Rightarrow S_n=3[1+2+3+.......\dots ]$

$S_n=3\cdot \sum N=\frac{3n(n+1)}{2}$

but we require

$S_n>1000$

$S_n=3\cdot \sum N=\frac{3n(n+1)}{2}>1000$

$n^2+n>\frac{2}{3}\times 1000$

$n^2+2.n.\frac{1}{2}+\frac{1}{4}>666.66+\frac{1}{4}$

${(n+\frac{1}{2})}^2>666.9$

$\Rightarrow n+\frac{1}{2}>25.8$

$n>25.8-0.5=25.3$

Since $n>25.3$ and n is an integer,

$\therefore$ the least value of n greater then $25.3$ is $26$.