Question

If the sum of the series $2+5+8+\mathrm{11.....}$ is $60100$, then the number of terms are

• A. $100$
• B. $200$
• C. $250$
• D. $300$

Answer 1

Answer: (B)

$200$

Given AP: $2+5+8+11+...$

Here, first term$,a=2$
Common difference$,d=3$

Given: Sum of terms of the AP $=60100$

We know that sum of terms of an AP $=\frac{n}{2}(2a+(n-1)\times d)$

$\Rightarrow \frac{n}{2}(2\times 2+(n-1)\times 3)=60100$

$\Rightarrow \frac{n}{2}(4+3n-3)=60100$

$\Rightarrow \frac{n}{2}(3n+1)=60100$

$\Rightarrow 3n^2+n=120200$

$\Rightarrow 3n^2+n-120200=0$

$\Rightarrow 3n^2+601n-600n-120200=0$

$\Rightarrow n(3n+601)-200(3n+601)=0$

$\Rightarrow (n-200)(3n+601)=0$

$\Rightarrow n=200$ or $n=\frac{-601}{3}$
Since $n$ cannot be negative, hence $n=200$