# Does my solution work by accident?

The task description is as follows (sorry for my poor translation - it's the best I could think of):

There are three numbers that sum up to 105. They are subsequent items in a growing geometric sequence (a, b, c). The first of those numbers is the first number, the second number is the sixth number and the third number is the twenty-sixth number in some arithmetic sequence. What are those numbers?

Now, my solution is as follows:

Those three numbers expressed in terms of the second sequence are $$(a, a + 5r, a + 25r)$$ where r is the difference of the arithmetic sequence. If so, then this is true (am I right?):

$$a + a + 5r + a + 25r = 105\\ a = 35 - 10r$$

Therefore the first sequence can be expressed as follows (again: am I right?): $$(35 - 10r, 35 - 5r, 35 + 15r)$$ . And here comes the most important question in this post: am I allowed to divide each of those items in the final sequence by a number so that my next calculations will deal with lower numbers? I'm asking because I did that and in the end I got the right values for a, b and c, although I'm not sure if I didn't do something wrong (prob. I did). So, my next calculations were:

$$(35 - 10r, 35 - 5r, 35 + 15r) / :5 \\ (7 - 2r, 7 - r, 7 + 3r)$$

From the formula for a second item of a geometric sequence ($$y^2 = x * z$$ where y is the second item in the sequence, x is the first one and z the last one):

$$(7 - r)^2 = (7 - 2r)(7 + 3r)\\ 49 - 14r + r^2 = 49 + 21r - 14r - 6r^2\\ 7r^2 = 21r\\ r = 3$$

Now if I put $$r$$ into $$(35 - 10r, 35 - 5r, 35 + 15r)$$, the three numbers come out as $$a = 5\\ b = 20\\ c = 80$$

Why does this work?

Looks correct. The reason it works to divide through is that the equation you write for requiring that $$a,b,c$$ are a geometric sequence only depend on the relative proportions of $$a,b,c$$. The equation you get is $$b/a = c/b$$ (equivalent to $$b^2 = ac$$), and you just reduced these fractions.