# Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?

I was wondering if anyone has any insights regarding the fact that the sum of any $a_1, \dots, a_{10}$ consecutive Fibonacci numbers is divisible by $11$ (and furthermore equals to $a_7*11$). What can this tell us about the series as a whole (its structure etc)?

Just write every term in the sum in terms of $a_1$ and $a_2$ (keeping in mind that $a_{n+2}=a_n+a_{n+1}$): $$a_1+a_2+(a_1+a_2)+(a_1 + 2a_2)+(2a_1+3a_2)+(3a_1+5a_2)+(5a_1+8a_2)+(8a_1+13a_2)+(13a_1+21a_2)+(21a_1+34a_2).$$
Then the sum is clearly equal to $55a_1+88a_2 = 11(5a_1+8a_2)$, which is $11$ times the seventh term of the sum.