Question

If the sum of n terms of the series 3, 6, 9, 12, ..., is 63000. Find the value of n.

• A. 300
• B. 200
• C. 150
• D. 100

Answer 1

The correct option is B 200

First term, a= 3
Common difference, d = 9-6 = 6-3 = 3
If first term is 'a' and common difference is 'd', sum of first 'n' terms $=\frac{n}{2}[2a+(n-1\left)d\right]$
Sum $=\frac{n}{2}[6+(n-1\left)3\right]=63000\Rightarrow \frac{n}{2}[3n+3]=63000\Rightarrow n^2+n-42000=0\Rightarrow n^2+210n-200n-42000=0\Rightarrow n(n+210)-200(n+210)=0\Rightarrow (n-200)(n+210)=0\Rightarrow n=200$