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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.
Thank you for your help!!!
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.
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
This should be a + c - a - b = 2(b + c - a - b).Quote:
Originally Posted by DarkAngle
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
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?Quote:
Originally Posted by DarkAngle
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
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.Quote:
Originally Posted by DarkAngle
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