# Geometric progression and arithmetic progression problem.

• Aug 9th 2012, 07:14 AM
DarkAngle
Geometric progression and arithmetic progression problem.
Hi,
I'm having difficulties with a problem with arithmetic and geometric progression i appreciate your help.
Here's the problem:
a,b,c form a geometric progression and a+b,b+c,a+c form an arithmetic progression.
I need to find the q=b/a=c/b.
• Aug 9th 2012, 07:37 AM
emakarov
Re: Geometric progression and arithmetic progression problem.
Express b and c through a and q, write the equation saying that the difference between the third and first terms of arithmetic progression is twice the difference between the second and first terms, simplify and solve for q. Don't forget special cases a = 0 and q = 1.
• Aug 9th 2012, 08:11 AM
DarkAngle
Re: Geometric progression and arithmetic progression problem.
b=a*q
c=a*q^2
(a+c-a+b)=2(b+c-a+b)
c+b=4b+c-a
3b=a
3aq=a 3q=a/a 3q=1 q=1/3 Is that correct?
Thank you for youe help!!!! :D
• Aug 9th 2012, 08:14 AM
emakarov
Re: Geometric progression and arithmetic progression problem.
Quote:

Originally Posted by DarkAngle
(a+c-a+b)=2(b+c-a+b)

This should be a + c - a - b = 2(b + c - a - b).
• Aug 9th 2012, 08:27 AM
DarkAngle
Re: Geometric progression and arithmetic progression problem.
Yes i forgot thanks for the correction
a+c-a-b=2(b+c-a-b)
c-b=2c-2a
-c-b=-2a ---> c+b=2a
aq^2+aq=2a
a(q^2+q)=2a ---> q^2+q-2=0 1-4*-2=9 ---> (-1+3)/2=1 (-1-3)/2=-2
seems this equation has two solution
• Aug 9th 2012, 08:32 AM
emakarov
Re: Geometric progression and arithmetic progression problem.
Quote:

Originally Posted by DarkAngle
(-1+3)/2=1 (-1-3)/2=-2
seems this equation has two solution

That's right. Check that a+b, b+c, a+c indeed form an arithmetic progression for each of those q. Also note that you divided both sides of a(q^2+q)=2a by a, which is possible only if a ≠ 0. What happens when a = 0?
• Aug 9th 2012, 09:13 AM
DarkAngle
Re: Geometric progression and arithmetic progression problem.
If a=0 then we don't have an equation at all.a,b,c is an geometric progression which mean a=0 leads to b=0 and c=0.
b=a*q a=0
b=0*q=0
c=a*q^2
c=0*q^2=0
• Aug 9th 2012, 09:15 AM
DarkAngle
Re: Geometric progression and arithmetic progression problem.
The first term must be different to 0 to have an geometric progression the fact that it is a geometric progression means a different to 0
• Aug 9th 2012, 09:25 AM
emakarov
Re: Geometric progression and arithmetic progression problem.
Quote:

Originally Posted by DarkAngle
The first term must be different to 0 to have an geometric progression the fact that it is a geometric progression means a different to 0

This depends on the definition, though Wikipedia, MathWorld and the Encyclopedia of Mathematics do not seem to require that a ≠ 0, only that q ≠ 0. If zero elements are not allowed, then dividing by a is justified and there are two solutions for q. If zero elements are allowed, then q can be any nonzero number.
• Aug 9th 2012, 09:33 AM
DarkAngle
Re: Geometric progression and arithmetic progression problem.
In my class and my book it says that every term must be different from 0
if a=0 then second term b=0
q=b/a ---> q=0/0 you can't divine by zero and that leads to nonexistence of q