# Arithmetic Series

• Sep 9th 2010, 10:04 PM
acc100jt
Arithmetic Series
Consider the Arithmetic progression with first term 1, common difference 3 and last term 100. If A is any set of 19 distinct numbers from the arithmetic progression, show that there must always be two numbers whose sum is 104.

Thank you!
• Sep 10th 2010, 12:35 AM
CaptainBlack
Quote:

Originally Posted by acc100jt
Consider the Arithmetic progression with first term 1, common difference 3 and last term 100. If A is any set of 19 distinct numbers from the arithmetic progression, show that there must always be two numbers whose sum is 104.

Thank you!

The series is $a_n=2n-2,\ n=1, .. , 34$. That the $n$ and $m$ th terms sum to $104$ means:

$(3n-2)+(3m-2)=104$

or:

$n+m=36$

Now both of $n=17$ and $m=19$ are in any set of $19$ consecutive integers between $1$ and $34$ inclusive, etc.. .

CB
• Sep 10th 2010, 01:32 AM
acc100jt
What happen if i pick 19 integers between 1 and 34 inclusive, but may not be consectutive, then n=17 and m=19 may not be in the set...hmm
• Sep 10th 2010, 02:34 AM
CaptainBlack
Quote:

Originally Posted by acc100jt
What happen if i pick 19 integers between 1 and 34 inclusive, but may not be consectutive, then n=17 and m=19 may not be in the set...hmm

Oppss.. misread the question, will get back to you on that

CB