# [SOLVED] a number series

• Nov 28th 2006, 09:39 PM
EcoDTR
[SOLVED] a number series
This is probably quite simple, but for some reason, I am having a problem with it.

Ok, I'm trying to find x in this series of numbers

1 , 1x , 1x^2 , 1x^3 , ..... , 0.2

the sum of all numbers in the series = 5

I would assume the solution has to do with logarithms.

If someone could show me how to find the solution to this, I would greatly appreciate it, thanks.
• Nov 28th 2006, 10:14 PM
topsquark
Quote:

Originally Posted by EcoDTR
This is probably quite simple, but for some reason, I am having a problem with it.

Ok, I'm trying to find x in this series of numbers

1 , 1x , 1x^2 , 1x^3 , ..... , 0.2

the sum of all numbers in the series = 5

I would assume the solution has to do with logarithms.

If someone could show me how to find the solution to this, I would greatly appreciate it, thanks.

This is apparently a finite geometric series, so we could apply the sum formula. My problem in solving this is that we don't know the number of elements in the sequence. Are you sure there isn't any more information?

-Dan
• Nov 28th 2006, 10:24 PM
topsquark
Quote:

Originally Posted by EcoDTR
This is probably quite simple, but for some reason, I am having a problem with it.

Ok, I'm trying to find x in this series of numbers

1 , 1x , 1x^2 , 1x^3 , ..... , 0.2

the sum of all numbers in the series = 5

I would assume the solution has to do with logarithms.

If someone could show me how to find the solution to this, I would greatly appreciate it, thanks.

I've been fiddling with this a bit. We know that the series is geometric, so
$\displaystyle a_n = a_0r^n$
where r = x and n is the number of terms in the sequence.

So we know that $\displaystyle a_n = 0.2$ for some integer n. Thus we can find that $\displaystyle x = (0.2)^{1/n}$ for some n.

We also have that
$\displaystyle S_n = 5 = \frac{1 - x^{n+1}}{1 - x}$

So plugging in various values of n we can get x and thus $\displaystyle S_n$. The problem is, no matter how large n is I can't get the sum to be larger than 1.25! (In fact 1.25 is the limit for $\displaystyle n \to \infty$.)

I don't know if I'm doing this wrong, or if the problem can't be done.

-Dan
• Nov 29th 2006, 04:24 AM
Soroban
Hello, EcoDTR!

That can't be the original problem.

Why all those 1's? . . . and what is that "0.2"? .It doesn't fit.

If the problem was actrually: .$\displaystyle 1 + x + x^2 + x^3 + \hdots \;=\;5$

. . we can solve for $\displaystyle x$ . . . otherwise, forget it!

• Nov 29th 2006, 05:09 AM
CaptainBlack
Quote:

Originally Posted by EcoDTR
This is probably quite simple, but for some reason, I am having a problem with it.

Ok, I'm trying to find x in this series of numbers

1 , 1x , 1x^2 , 1x^3 , ..... , 0.2

the sum of all numbers in the series = 5

I would assume the solution has to do with logarithms.

If someone could show me how to find the solution to this, I would greatly appreciate it, thanks.

The information you give is that the sequence is:

$\displaystyle 1,\ x,\ x^2,\ ...,\ x^n$

for some $\displaystyle n$, and that $\displaystyle x^n=0.2$, and that:

$\displaystyle 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}=5$

So we have:

$\displaystyle \frac{1-0.2 \times x }{1-x}=5$

which has solution $\displaystyle x=5/6$.

But we also have $\displaystyle x^n=0.2$, so:

$\displaystyle n \log(x)=\log(0.2)$

or:

$\displaystyle n=\frac{\log(0.2)}{\log(5/6)} \approx 8.8$

which is not an integer, which contradicts our assumptions, so there is no solution to the problem as stated.

RonL