Attachment 29534

(Itwasntme)

Printable View

- Oct 21st 2013, 07:46 AMparmisGeometric progression
Attachment 29534

(Itwasntme) - Oct 21st 2013, 10:36 AMtopsquarkRe: Geometric progression
- Oct 22nd 2013, 03:28 AMparmisRe: Geometric progression
Oh... sorry i made a mistake. a n is Arithmetic progression! d=1/5

- Oct 22nd 2013, 05:09 AMHallsofIvyRe: Geometric progression
So $\displaystyle a_n$ is an

**arithmetic**progression of the form $\displaystyle a_n= a+ nr$ but what does "m" represent in your formula? - Oct 22nd 2013, 08:10 AMparmisRe: Geometric progression
I dont know. My teacher gave me this, but i cant answer it!

- Oct 22nd 2013, 09:21 AMSlipEternalRe: Geometric progression
Then you should probably ask your teacher what $\displaystyle m$ means.

- Oct 22nd 2013, 09:34 AMSlipEternalRe: Geometric progression
Assuming your arithmetic progression is $\displaystyle a_n = a+dn$ since you arbitrarily wrote a d for some reason, we can set these two expressions equal to each other:

$\displaystyle a_n = a+dn$

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

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

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

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