Rhind Papyrus 40

Problem

 * A hundred loaves to five men, one-seventh of the three first men to the two last.

Solution
This means:
 * Divide $100$ loaves between $5$ men so that:
 * the shares are in arithmetic progression
 * the sum of the two smaller shares is $\dfrac 1 7$ of the sum of the $3$ greatest.

Thus, let:
 * $a$ be the smallest share, that is, the initial term
 * $d$ be the common difference.

We have:

So the shares are: