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annar
Posts: 1
Joined: Sat Sep 17, 2016 8:39 am

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

aritra barua
Posts: 50
Joined: Sun Dec 11, 2016 2:01 pm

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...

Posts: 90
Joined: Sun Jan 28, 2018 11:43 pm
Location: Bhulta, Rupganj, Narayanganj

How have you solve the problem? Can someone suggest me?
যদি জীবনের গোড়ায় কষ্ট কর, তবে পরবর্তী জীবনে কম কষ্ট করতে হবে।

samiul_samin
Posts: 321
Joined: Sat Dec 09, 2017 1:32 pm
Location: Shantinagar,Digharkanda,Mymensingh
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Sat Feb 17, 2018 9:39 pm
How have you solve the problem? Can someone suggest me?
Solution
$1+2+3+4+5+6+7+8+9+10+11+12=78$
$\int^{\infty}_0 e^{-x} dx=1$

Posts: 90
Joined: Sun Jan 28, 2018 11:43 pm
Location: Bhulta, Rupganj, Narayanganj

samiul_samin wrote:
Sat Feb 17, 2018 10:43 pm
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)
যদি জীবনের গোড়ায় কষ্ট কর, তবে পরবর্তী জীবনে কম কষ্ট করতে হবে।

samiul_samin
Posts: 321
Joined: Sat Dec 09, 2017 1:32 pm
Location: Shantinagar,Digharkanda,Mymensingh
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Here is the Rule
$1+2+3+... ... ...+n=n(n+1)/2$
It can be asily proved by Induction
$\int^{\infty}_0 e^{-x} dx=1$

Posts: 90
Joined: Sun Jan 28, 2018 11:43 pm
Location: Bhulta, Rupganj, Narayanganj

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
যদি জীবনের গোড়ায় কষ্ট কর, তবে পরবর্তী জীবনে কম কষ্ট করতে হবে।

samiul_samin
Posts: 321
Joined: Sat Dec 09, 2017 1:32 pm
Location: Shantinagar,Digharkanda,Mymensingh
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Hint
Use Middle term factor
$n(n+1)=156\Rightarrow n^2+n-156=0 \Rightarrow n^2-12n+13n-156=0 \Rightarrow n(n-12)+13(n-12)=0 \Rightarrow (n-12)(n+13)=0$
$\int^{\infty}_0 e^{-x} dx=1$

Posts: 90
Joined: Sun Jan 28, 2018 11:43 pm
Location: Bhulta, Rupganj, Narayanganj

samiul_samin wrote:
Sun Feb 18, 2018 10:26 am
Hint
Use Middle term factor
$n(n+1)=156\Rightarrow n^2+n-156=0 \Rightarrow n^2-12n+13n-156=0 \Rightarrow n(n-12)+13(n-12)=0 \Rightarrow (n-12)(n+13)=0$
How can I get the rate of 'n'
যদি জীবনের গোড়ায় কষ্ট কর, তবে পরবর্তী জীবনে কম কষ্ট করতে হবে।

samiul_samin
Posts: 321
Joined: Sat Dec 09, 2017 1:32 pm
Location: Shantinagar,Digharkanda,Mymensingh
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$n(n+1)=156\Rightarrow n^2+n-156=0 \Rightarrow n^2-12n+13n-156=0 \Rightarrow n(n-12)+13(n-12)=0 \Rightarrow (n-12)(n+13)=0$
$n=12,-13,n>0$ so,$n=12$
$\int^{\infty}_0 e^{-x} dx=1$