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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.