Greek Anthology Book XIV: 12. - Problem

Problem

 * Croesus the king dedicated six bowls weighing six ,
 * each one heavier than the other.

Solution
It is known that there are $100$ to the.

The weights of the bowls form an arithmetic progression.

Let $a$ denote the weight of the lightest bowl in.

Let $n$ denote the number of bowls.

Let $d$ denote the common difference between the weights in of consecutive bowls arranged in order of weight.

Let $t$ denote the total weight of all bowls in.

We are given:

So:

So the lightest bowl weighs $97 \frac 1 2$, and it follows that the weights of each bowl in is:


 * $97 \frac 1 2, 98 \frac 1 2, 99 \frac 1 2, 100 \frac 1 2, 101 \frac 1 2, 102 \frac 1 2$