Sequence of 100 real numbers

We have a sequence of real numbers $a_1,a_2,\dots,a_{100}$ such that the average of every 7 terms is equal to average of some 11 terms. Show that $a_1=\cdots a_{100}$

Have tried this with pigeon hole and divisibility , only thing I got was that all the number's would have same remainder when divided by 7(r1) and 11(r2).If i do the same thing again and again using $a_1 =7k_1+r , a_2=7k_2+r, \ldots$ I again get $k_1$ to be divisible by 7 but how do I use this . How do I prove that they are equal

• Start by considering the average of the smallest 7 numbers. P.S. Since it's a sequence of real numbers (not necessarily integers) divisibility doesn't apply. – dxiv Aug 21 '16 at 17:23
• Hint: Wolog $a_1 \le a_2 \etc...$. You can prove average of smallest 7 is average of smallest 11 so average of a_8 to a_1 is the same average. – fleablood Aug 21 '16 at 18:19
• yeah but that proves first 11 number's to be same we cannot say that for all 100 numbers but yeah i did not think this before let me think a bit more on the same lines – Kashish Garg Aug 21 '16 at 19:04
• I cannot proceed with this . I know smallest 11 terms are equal and largest 11 terms are equal . But i cannot think about anything after that – Kashish Garg Aug 22 '16 at 11:26
• "average of every 7 terms" means "of every consecutive 7 terms" or "of every 7 terms whatever" ? – G Cab Aug 25 '16 at 8:42

A start, but I didn't reach the contradiction I expected. There must be an easy way to show you keep adding points indefinitely if two of the $a$'s have different values.
You say in the comments that you can show the smallest eleven are the same and the largest eleven are the same. If we have a group of $a$'s that work, we ca add or subtract a constant from all of them and multiply by another constant so the smallest number is $0$ and the next smallest number is $1$. Consider six of the smallest and the $1$. Their average is $\frac 17$ If your group of $11$ has two terms not equal to $0$ that are at least $1$ the average will be too high, if it has only one it must be $\frac {11}7$, which must be in the set. Now we can take $\frac {11}7$ with six of the zeros and get an average of $\frac {11}{49}$. The sum of the eleven that match this average is $\frac {121}{49}.$ If we have two numbers in the group of eleven with this average greater than $0$, neither can be $\frac {11}7$ as the other would be less than $1$.