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!
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