finding the value of $f(\frac{1}{7})$ $f$ is a function mapping positive reals between $0$ and $1$ to reals. Let $f$ be given by, $f( \frac{x+y}{2} ) = (1-a)f(x)+af(y)$ where $y > x$ and $a$ being a constant. Also,$f(0) = 0$ and $f(1) = 1$. Find $f( \frac{1}{7} )$.
I have tried some things but it is not working out. How to find the $f( \frac{1}{7} )$ . 

For each $n$ with $1\leq n\leq 6$ we have the equation $f(\frac{n}{7}) = (1-a)f(\frac{n-1}{7}) + af(\frac{n+1}{7})$.  Since we know $f(0)$ and $f(1)$, this gives us a system of $6$ linear equations in $6$ unknowns.
