# Arithmetic Series

• Sep 9th 2010, 09:04 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 9th 2010, 11:35 PM
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 \$\displaystyle a_n=2n-2,\ n=1, .. , 34\$. That the \$\displaystyle n\$ and \$\displaystyle \$\$ m\$ th terms sum to \$\displaystyle 104\$ means:

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

or:

\$\displaystyle n+m=36\$

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

CB
• Sep 10th 2010, 12:32 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, 01:34 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