One person enters the stadium the first mimute, two people the second mimute, three people the third minute, and so on. In how many mimutes will there be 55 people in the stadium?

- Jun 22nd 2006, 08:27 PMmaggieI am a newbie. Please help me to solve this problem. I would appreciate your help.
One person enters the stadium the first mimute, two people the second mimute, three people the third minute, and so on. In how many mimutes will there be 55 people in the stadium?

- Jun 22nd 2006, 08:40 PMearbothQuote:

Originally Posted by**maggie**

you want to know the number of summands in a sum:

$\displaystyle 1+2+3+...+n=55\Longrightarrow \frac{1}{2}\cdot n\cdot (n+1)=55$

Expand the LHS of this equation and solve the quadratic equation for n:

$\displaystyle \frac{1}{2}n^2+\frac{1}{2}n-55=0$

You'll get x = -11 or x = 10. (-11) persons are a very ugly sight.

So the only possible solution for your problem is: 10 persons.

Greetings

EB - Jun 22nd 2006, 08:42 PMCaptainBlackOther copy of this deleted
Don't make multiple posts of the same question.

RonL - Jun 22nd 2006, 09:15 PMmaggie
Thank you for your help. However, my son is 11 years old. I don't think he could understand the fomular you just gave. Do you have another way to show my son to understand easier. Thank you so much.

- Jun 22nd 2006, 10:20 PMearbothQuote:

Originally Posted by**maggie**

show your son how to calculate in a table.

I've attached a diagram to show you what I mean.

Greetings

EB - Jun 23rd 2006, 09:13 AMmaggieHello Earboth,
Thank you so much for your help. My son is understanding now. It is much easier for him to understand . Thank you again.