A 66-seat bus picks up 1 passenger at the 1st bus stop, 2 passengers at the 2nd stop, 3 passengers at the 3rd bus stop and so on. When will all the seats in the bus be filled?

Guest Apr 6, 2021

#1

#3**+2 **

How did you get 11? See my solution for how I got 12. I think you might have factored the quadratic the wrong way as $(n+12)(n-11)=0$ or something.

RiemannIntegralzzz
Apr 6, 2021

#4**+2 **

really? i honestly just used the long way and just added until i got 66, 1+2+3+4+5+6+7+8+9+10+11 is 66, i believe. i may have done it wrong, sorry!

Guest Apr 6, 2021

#5**+2 **

Wait you were right, I just realized that I actually factored the quadratic the wrong way :)) the irony. Thank you for helping out!

RiemannIntegralzzz
Apr 6, 2021

#2**+2 **

You have the sequence $1+2+3+4+5+...+(n-1)+n$, so the sum of this will be $\frac{n(n+1)}{2}$.

So, you are looking for $\frac{n(n+1)}{2}=66$.

$n(n+1)=132$

$n^2+n-132=0$

$(n+12)(n-11)=0$.

This means that $n=11$ or $n=-12$. $n$ obviously cannot be negative, so the only solution is $n=11$, so at the $\boxed{11^{th}}$ stop.

RiemannIntegralzzz Apr 6, 2021