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Hey guys,
I am no maths genius and need help with this itsy bitsy problem. I know I can easily find the solution by adding them one by one but I was wondering if someone can come up with a quicker solution to this problem, some sort of a formula I guess.

This is the problem:

A train will hold 78 passengers. If it starts out empty and picks up one passenger at the first stop, two passengers at the second stop, 3 passengers at the third stop and so on. After how many stops will it be full?

Thanks, appreciate the help

Well,this simply reduces the problem to the sum of n consecutive positive integers,which follows the equation :n(n+1)=156,simply factoring the expression,we find n=12 and n cannot be -13 as it is a negative integer.This just solves our problem that at the 12th stop,the bus would be full...

How have you solve the problem? Can someone suggest me?

M. M. Fahad Joy wrote: ↑

Sat Feb 17, 2018 9:39 pm

How have you solve the problem? Can someone suggest me?

Solution

samiul_samin wrote: ↑

Sat Feb 17, 2018 10:43 pm

M. M. Fahad Joy wrote: ↑

Sat Feb 17, 2018 9:39 pm

How have you solve the problem? Can someone suggest me?

Solution

$1+2+3+4+5+6+7+8+9+10+11+12=78$

No, I said by the rules on N(N+1)

Here is the Rule

samiul_samin wrote: ↑

Sun Feb 18, 2018 9:24 am

Here is the Rule

$1+2+3+... ... ...+n=n(n+1)/2$
It can be asily proved by Induction

Thanks, understood. But how do you solve the equation?

n(n+1) = 156

Hint Answer

samiul_samin wrote: ↑

Sun Feb 18, 2018 10:26 am

Hint

Use Middle term factor

Answer

$n(n+1)=156\Rightarrow n^2+n-156=0
\Rightarrow n^2-12n+13n-156=0
\Rightarrow n(n-12)+13(n-12)=0
\Rightarrow (n-12)(n+13)=0$

How can I get the rate of 'n'

M. M. Fahad Joy wrote: ↑

Sun Feb 18, 2018 10:29 am

samiul_samin wrote: ↑

Sun Feb 18, 2018 10:26 am

Hint

Use Middle term factor

Answer

$n(n+1)=156\Rightarrow n^2+n-156=0
\Rightarrow n^2-12n+13n-156=0
\Rightarrow n(n-12)+13(n-12)=0
\Rightarrow (n-12)(n+13)=0$

How can I get the rate of 'n'

$n=12,-13,n>0$ so,$n=12$