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Good evening, I've been struggling with this one for a while now.

The sum of an increasing geometrical progression is equal to 65. If we substract 1 from the smallest number there, and 19 from the biggest, the three numbers are going to form an arithmetical progression. Find these numbers.

I've gone trough many methods and none of them worked, so I't would be nice if you could explain this step by step. Not to look like someone just hunting for answers, I'll provide what I know:
The geometric progression, consisted of members b1, b2, b3. The sum of it is equal to 65. b1q = b2; b1q^2=b3

The aritmetic progression, consisted of a1, a2, a3. a1 = b1 - 1; a2 = b2 ; a3 = b3 - 19. a1 + d = a2, a1 + 2d = a3 I'm sort of new to this site, so sorry if I didnt provide something, like tags, because these type of things are called progressions here and there no such things in tag list. Im in 12th grade which is equal to senior year in USA as far as I know if that matters. thank you!

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    $\begingroup$ First off, $b_2=\sqrt{b_1b_3}$ - also, you need to prove that it's only 3 terms... $\endgroup$ – Shakespeare Oct 2 '14 at 18:54
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    $\begingroup$ the sum of a geometric progression has a very well known explicit form... $\endgroup$ – Alex R. Oct 2 '14 at 18:54
  • $\begingroup$ oh ok. And what would you recommend doing next ? I meant this about the arithmetic progression then . $\endgroup$ – Benskey Oct 2 '14 at 18:56
  • $\begingroup$ can I say that a2 = sqrt(b1b3) ? $\endgroup$ – Benskey Oct 2 '14 at 18:58
  • $\begingroup$ Yes, that will be fine. I would suggest not using subscripts, they can be a source of confusion. Symbol manipulation will do the job, but there are simpler ways. $\endgroup$ – André Nicolas Oct 2 '14 at 19:00
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You can develop a system of two equations in two unknowns as follows. Represent the terms of the geometric progression as $c$, $rc$, and $r^2c$. Since you are told that they sum to $65$, you know $$c + rc + r^2c = 65\tag{1}$$ Now the terms of the arithmetic progression must be $c-1$, $rc$, and $r^2c-19$. Since this is an arithmetic progression, the difference between any two successive terms is the same; thus $$(rc)-(c-1) = (r^2c - 19)-(rc)\tag{2}$$ Equations (1) and (2) provide your system of equations. Cleaning them up a little bit, you have the system $$\left\{\begin{array}{l}r^2c + rc + c = 65 \\ r^2c - 2rc + c = 20 \end{array}\right.$$ to solve for $r$ and $c$, which will then give you the terms of the progressions.

Subtracting the equations in this system gives you immediately that $3rc=45$. Since necessarily $r\neq 0$ (why?), you can substitute $c=15/r$ in (say) the first equation and clean it up to get $$3r^2 - 10r + 3 = 0$$ This factors as $$(r-3)(3r-1)=0$$ so the solutions are $r=3$ (which gives $c=5$) and $r=1/3$ (which gives $c=45$). Since the geometric progression is known to be increasing, we eliminate the solution $r=1/3$ (it produces a decreasing geometric progression). This means the geometric progression is $$\boxed{5,15,45}$$ and the arithmetic progression is $$\boxed{4,15,26}.$$

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  • $\begingroup$ I love you! Thanks!! $\endgroup$ – Benskey Oct 2 '14 at 20:40
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Here's some thoughts:

Your geometric progression could be $a, ar, ar^2$, with sum $65$. Also $a-1, ar, ar^2-19$ is an arithmetic progression with sum $45$. Hence the middle number, $ar$ is $15$.

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  • $\begingroup$ How did you get that the middle number, ar, is 15 ?Edit; I understood this, its because you substract 1+19. $\endgroup$ – Benskey Oct 2 '14 at 18:59
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    $\begingroup$ In an arithmetic progression of length $3$, say $x, x+y, x+2y$, the middle number is the average of all three values. $\endgroup$ – paw88789 Oct 2 '14 at 19:02
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    $\begingroup$ @Benskey The geometric sequence sums to 65. Since we subtracted 1 and 18, we know the arithmetic sequence sums to 45. The middle number is the average of the sequence, and hence it is $45\div3=15$. $\endgroup$ – Akiva Weinberger Oct 2 '14 at 19:03
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First, $$b_1+b_2+b_3=65\iff b_1+b_1q+b_1q^2=65.$$ Second, $$a_1+a_3=2a_2\iff (b_1-1)+(b_3-19)=2b_2\iff (b_1-1)+(b_1q^2-19)=2b_1q.$$ Then, solving these for $b_1,q$ gives you $$(b_1,q)=(5,3),\left(45,\frac 13\right).$$ Hence, the answer is the former.

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  • $\begingroup$ Thank you. Could you elaborate on the solving for b1 and q ? Im a little confused there! I got that b1(q^2 - 2q + 1) = 20, is this a step in the right direction ? $\endgroup$ – Benskey Oct 2 '14 at 19:12
  • $\begingroup$ @Benskey: Well, how about this? From the second, $b_1+b_1q^2=2b_1q+20$. So, from the first, $3b_1q+20=65\Rightarrow b_1q=15\Rightarrow q=15/b_1$. Then, from the first, you can get $b_1^2-50b_1+225=0\Rightarrow (b_1-5)(b_1-45)=0$. $\endgroup$ – mathlove Oct 2 '14 at 19:15

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