Finding x in a geometric progression, given the sum.

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NotaPhysicist
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Homework Statement



If

[tex]1 + 2x + 4x^2 + ... = \frac{3}{4}[/tex]

find the value of x. [Edit: Forgot to ask the question]

Homework Equations



[tex]S_n = \frac{a(1 - r^n)}{1 - r}[/tex]

[tex]t_n = ar^{n-1}[/tex]

The Attempt at a Solution



a = 1

r = 2x

I try to solve [tex]S_n[/tex] and end up with

[tex]2x^n = \frac{6x - 7}{4}[/tex]

which I can't solve.

I try to solve by equating t2 and t3 and getting x = (1/2). Which is wrong.

Any help appreciated.
 
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Is the sum is to infinity?
or to 'n terms'?

If it is to infinity, Apply limit to your equation.
 
NotaPhysicist said:

Homework Statement



If

[tex]1 + 2x + 4x^2 + ... = \frac{3}{4}[/tex]


Homework Equations



[tex]S_n = \frac{a(1 - r^n)}{1 - r}[/tex]
If this is an infinite sum the formula is
[tex]S_\infty= \frac{a}{1- r}[/tex]

If it is a finite sum, you would need to know how many terms so that "n" would be an actual integer, not a variable.

The "..." at the end of the sum indicates it is an infinite sum.

[tex]t_n = ar^{n-1}[/tex]

The Attempt at a Solution



a = 1

r = 2x

I try to solve [tex]S_n[/tex] and end up with

[tex]2x^n = \frac{6x - 7}{4}[/tex]

which I can't solve.

I try to solve by equating t2 and t3 and getting x = (1/2). Which is wrong.

Any help appreciated.
 
I've edited the original post. The problem is to find the value of x.

Its not an infinite sum. The only solution I can see is to solve for n in the infinite series and the summation, and try to solve simultaneously.
 
HallsofIvy said:
The "..." at the end of the sum indicates it is an infinite sum.

I had to read your answer a couple of times. I get it now. Its an infinite sum. No powers to work out.

Thank you so much.