In a jar of 115 balls, each being red or blue. There are 60 red balls. Show that there are at least two red balls in the jar exactly 4 terms apart.
My Approach
There are 115 balls and we need 4 terms apart, so we make groups of 6
$115 \over6$$ = 19.167$
also we know we have 60 red balls $\therefore$ $60 \over6$$ = 10$
Since we have 2 possibilities, red or blue from pigeon holes principle, we can say that red and blue both have $115 \over(6*2)$$= $$115 \over12$ which is approximately $9.583$ , which means that out of red and blue, one has $10$ pairs and the other has $9$.
Also we know that there are 10 pairs red balls.
This shows that it is true that the red balls have 10 pairs of 6 balls which are red. This also shows that in the pair of 6 balls, the first and the last is also red and they are 4 terms apart in turn proving the hypothesis.
I am not sure if I did it right as this seems too sketchy
It does seem rather sketchy ... in fact, I can't follow it at all. Groups of $4$? red and blue? Pair of $6$ balls?
Also, the question itself is weird: balls in a jar ... 4 terms apart?
OK, I assume that the question basically is: If you pick $60$ numbers from $1$ through $115$, then there must be two numbers with a difference of $4$
Here's why:
Pick $60$ numbers.
Divide them into $4$ groups, depending on whether the number $\equiv 0,1,2$, or $3 \bmod 4$
Note that in each group, the numbers are a multiple of $4$ apart.
So, if you have a group with at least $16$ numbers, then for no two of them to be exactly $4$ apart, they need to be at elast $8$ apart, but that means the lowest and highest number are at least $8 \cdot 15 =120$ apart, which is impossible.
So, all $4$ groups have exactly $15$ members. But now you have a problem with the group of numbers that are $\equiv 0 \bmod 4$, for the lowest possible number is $4$, and the highest needs to be at least $8 \cdot 14=112$ higher, which is $116$, and so there must be at least two that are exactly $4$ apart.