Proving the Convergence of a Geometric Series with a Tricky Sum Equation

[misogynisticfeminist]

I've got a problem here...

A geometric series has first term 1,the sum of the first 5 terms is twice that of the sum of the 6th to 15th term inclusive. Prove that r^5= \frac{1}{2} \sqrt {3-1}

What i did was...

2s_5=s_{15}-s_5

using the formula for the sum of a GS, i got...

2r^4 -1 =r^{14} -r^4

and things when downhill from there.

I've tried expressing it in terms of x^5 because the answer seems to suggest the quadratic formula. But i don't seem to be getting anywhere. Can anyone help?

 

[VietDao29]

Hmm, it may be me with a wrong calculation, or is it the answer wrong?
Anyway, I'll give you a hint:
A_1 = 1
A_n = r ^ {n - 1}
So A_6 = ?
If you assume A_6 the first term of another geometric series, with the same r, can you find the sum of the first 10 terms (6th - 15th)?
Viet dao,

 

[StatusX]

A few things. Is that 3-1? Or, as I like to call it, 2? Also, the formula for the sum of a geometric series, from r^n to r^m is (r^(m+1) - r^n)/(r-1). And finally, didn't you say the sum of first set was twice the sum of the second? Because it looks like you have that backwards in your equation.

 

[HallsofIvy]

The first term is one so this series is just 1+ r+ r²+ r³+ ... for some r.
The sum of the first five terms is \frac{1-r^6}{1-r} and the sum of the "6th to 15th term" is "sum of first 15 terms minus sum of first 5 terms":
\frac{1-r^{16}}{1-r}-\frac{1-r^6}{1-r}= \frac{r^6-r^16}{1-r}.
We are told that "sum of the first 5 terms is twice that of the sum of the 6th to 15th term" so solve the equation \frac{1-r^6}{1-r}= \frac{r^6-r^16}{1-r}. Solve that for r⁵.

 

[StatusX]

I think those powers should be 5 and 15. The fifth term of the series is r^4. And the actual answer has the 1 outside the radical, but still multiplied by 1/2.

 

[VietDao29]

Isn't this the sum of the first 5 terms:
S_5 = 1 + r + r^2 + r^3 + r^4 = \frac{r^5 - 1}{r - 1}
Or this:
S_5 = 1 + r + r^2 + r^3 + r^4 + r^5= \frac{r^6 - 1}{r - 1}?
Viet Dao,

 

[OlderDan]

I think those powers should be 5 and 15. The fifth term of the series is r^4. And the actual answer has the 1 outside the radical, but still multiplied by 1/2.

I agree with this. Is the negative solution invalid? I don't see why it would be. There is no reason why the series has to converge for infinite n.

<br /> r^5 = \frac{{ \pm \sqrt 3 - 1}}{2} \ \ ??<br />