Arithmetic progression/algebra problem

[NotaPhysicist]

Homework Statement

The three number a, b, c, whose sum is 15, are successive terms of an geometric sequence, and b, a and c are successive terms of an arithmetic sequence. Find the values of a, b and c.

Edit: I wrote the question wrong. It now reads correctly. I also fixed my latex code.

Homework Equations

Geometric sequence:

$$t_n = ar^{n-1}$$

$$S_n = \frac{a(r^n - 1)}{r - 1}$$

Arithmetic sequence:

$$t_n = a + (n - 1)d$$

$$S_n = \frac{n}{2} \[2a + (n - 1)d\]$$

The Attempt at a Solution

I'm more or less baffled. I get lost in the algebra. Looking for a heads up on which path to follow.

And sorry about the latex code. As much as I try I can't get it right.

 

[Mark44]

Homework Statement

The three number a, b, c, whose sum is 15, are successive terms of an geometric sequence, and b, a and c are successive terms a geometric sequence. Find the values of a, b and c.

Homework Equations

Geometric sequence:

Fixed your LaTeX. The closing tag should be [ /tex] (without the leading space), not [\tex]. Also, if an exponent has more than one character, use braces around all characters in the exponent, like this, ^{n - 1}

$$t_n = ar^{n-1}\\ S_n = \frac{a(r^n - 1)}{r - 1}$$

Arithmetic sequence:

$$t_n = a + (n - 1)d\\ S_n = (n/2) [2a + (n - 1)d]$$

The Attempt at a Solution

I'm more or less baffled. I get lost in the algebra. Looking for a heads up on which path to follow.

And sorry about the latex code. As much as I try I can't get it right.

Set up your equations in a, b, and c.

For the first progression, you have a, a + d, and a + 2d as successive terms in the arithmetic progression, where d is the common difference. You know what they should add to.

For the second, you have that b, a, and c are successive terms in a geometric progression. If b is the first term in such a progression, what does a have to be? What does c have to be?

 

[sjb-2812]

Can you check your typed question, as you refer to both the order a,b,c and b,a,c as geometric sequences, so why you've mentioned the formula for an arithmetic, I'm not sure. Otherwise, I agree with Mark44's approach

 

[NotaPhysicist]

Sorry to everyone who tried to help. I have now retyped the original question.

a + b + c = 15 (successive terms of geometric progression)

b, a, c (successive terms of an arithmetic progression)

 

[HallsofIvy]

Sorry to everyone who tried to help. I have now retyped the original question.

a + b + c = 15 (successive terms of geometric progression)

b, a, c (successive terms of an arithmetic progression)

This is still unclear. Are a, b, and c the same as in the geometric progression? This is one problem?

If a, b, and c are succesive terms of a geometric progression then $b= ar$ and $c= ar^2$ for some number r. We have $a+ ar+ ar^2= a(1+ r+ r^2)= 15$. If b, a, and c, in that order, are terms in an arithmetic sequence we must have a= b+ d and c= b+ 2d for some "common difference" d. W can also write that as b= a- d and c= a- d+ 2d= a+ 2d. Their sum is now (a- d)+ (a)+ (a+ d)= 3a= 15 so a must be 5. That means $1+ r+ r^2= 3$ so that $r^2+ r- 2= (r+ 2)(r- 1)= 0$.

Now do either of those give the sequence a, b, c so that a, b, c is a geometric sequenc and b, a, c is an arithmetic sequence?

 

[NotaPhysicist]

Yes, this is the same problem, word for word from the text.

Unfortunately I can't follow your progress. I can follow up until

(a- d)+ (a)+ (a+ d)= 3a= 15

I don't know how you got that. I've fooled around with it a bit, I won't embarrass myself to say exactly how long I've spent trying to solve this. This is senior high school level stuff.

I may have to pay the $33 per chapter for worked solutions some guy is offering over the Internet..

Thanks anyway.

 

[sjb-2812]

Yes, this is the same problem, word for word from the text.

Unfortunately I can't follow your progress. I can follow up until

(a- d)+ (a)+ (a+ d)= 3a= 15

I don't know how you got that.

How do you define an arithmetic sequence?

 

[NotaPhysicist]

I get it now. Since is the second term in the arithmetic sequence

(a- d)+ (a)+ (a+ d)= 3a= 15
a= 3

I thought that would lead to a solution. But I'm still stuck. Too many pages of wasted algebra. I'm giving up on this one.

Thanks again for your help.

 

[sjb-2812]

If 3a = 15, a <> 3 :)