# Geometric progression

• Oct 21st 2013, 07:46 AM
parmis
Geometric progression
Attachment 29534
(Itwasntme)
• Oct 21st 2013, 10:36 AM
topsquark
Re: Geometric progression
Quote:

Originally Posted by parmis
Attachment 29534
(Itwasntme)

I'm going to take a guess here. Is this your problem?

$a_n = \frac{(m^2 - a)n^2 +m + n}{m^2 + 2m + 1}$

But that can't be right...the a on the RHS needs a subscript, doesn't it?

-Dan
• Oct 22nd 2013, 03:28 AM
parmis
Re: Geometric progression
Oh... sorry i made a mistake. a n is Arithmetic progression! d=1/5
• Oct 22nd 2013, 05:09 AM
HallsofIvy
Re: Geometric progression
So $a_n$ is an arithmetic progression of the form $a_n= a+ nr$ but what does "m" represent in your formula?
• Oct 22nd 2013, 08:10 AM
parmis
Re: Geometric progression
I dont know. My teacher gave me this, but i cant answer it!
• Oct 22nd 2013, 09:21 AM
SlipEternal
Re: Geometric progression
Then you should probably ask your teacher what $m$ means.
• Oct 22nd 2013, 09:34 AM
SlipEternal
Re: Geometric progression
Quote:

Originally Posted by parmis
Oh... sorry i made a mistake. a n is Arithmetic progression! d=1/5

Assuming your arithmetic progression is $a_n = a+dn$ since you arbitrarily wrote a d for some reason, we can set these two expressions equal to each other:

$a_n = a+dn$

$= a+\dfrac{n}{5} = \dfrac{(m^2-a)n^2+m+n}{m^2+2m+1}$

If you are trying to solve for $m$, it is the solution to this quadratic equation in $m$:

$\left(a+\dfrac{n}{5}-n^2\right)m^2 + \left(2a + \dfrac{2n}{5}-1\right)m + \left(a+an^2-\dfrac{4n}{5}\right) = 0$

You can use the quadratic formula. You get a mess, but I suppose that could be the solution for $m$...