Question

If $2+5+8+11+\cdots \text{upto}n\text{terms}=610$, then the value of $n$ is

• A. $19$
• B. $22$
• C. $21$
• D. $20$

Answer 1

The correct option is D $20$

Given: $2+5+8+\cdots \text{upto}n\text{terms}=610$
The given sequence is an A.P. with $a=2,d=3$
Now, sum of $n$ terms of A.P. is
$=\frac{n}{2}[4+3(n-1\left)\right]=610\Rightarrow 4n+3n(n-1)=1220\Rightarrow 3n^2+n-1220=0\Rightarrow 3n^2+61n-60n-1220=0\Rightarrow (n-20)(3n+61)=0\therefore n=20(\because n>0)$