# A geometric series question

#### 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?

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#### VietDao29

Homework Helper
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,

Last edited:

#### StatusX

Homework Helper
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

Homework Helper
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

Homework Helper
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

Homework Helper
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

Homework Helper
StatusX said:
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.

$$r^5 = \frac{{ \pm \sqrt 3 - 1}}{2} \ \ ??$$

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