# Arithmetic and Geometric sequence problem

## Homework Statement

The sum of first three numbers of the arithmetic sequence is 54. If you subtract 3 from the first one, leave the second one unchanged and add 12 to the third one you get the first three numbers of the geometric sequence of the form $ar + ar^2 + ar^3 + ... ar^n$ Find r.

## Homework Equations

3. The Attempt at a Solution [/B]
Using the first clue i got $a + d = 18$ $a$ being the first number in either sequence and $d$ being the difference of the arithmetic sequence. Next i set up the geometric sequence of the given form $(a - 3) + (a + d) + (a + 2d + 12)$ . Now $\frac{a + d}{a - 3} = r$ and from this $\frac{18}{18 - d} = r$ . Also $\frac{a + 2d + 12}{a + d} = r^2$ . Now defining $18 - \frac{18}{r} = d$ and replacing in the formula above i obtain $8r - 3 = r^3$ . This does not hold for $r = 2$ which is the correct solution. Which step did i do wrong?

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Samy_A
Homework Helper
When you went from $\frac{a + d}{a - 3} = r$ to $\frac{18}{18 - d} = r$, you forgot the $-3$ of the first denominator.

When you went from $\frac{a + d}{a - 3} = r$ to $\frac{18}{18 - d} = r$, you forgot the $-3$ of the first denominator.
True but didnt make much difference. i now get $5r - 2 = 2r^3$ Cant find any mistakes but there must be some.

Samy_A
$\frac{a + 2d + 12}{a + d} = r^2$ is not correct.
• yeah, thats it. Just r there. Now its correct $5r - 2 = 2r^2$