# Homework Help: Geometric Sequences & Series

1. Jun 11, 2010

### wvcaudill2

1. The problem statement, all variables and given/known data
The sum of the first six terms in a geometric sequence of real numbers is 252.
Find the sum of the first four terms when the sum of the first two terms is 12.

2. Relevant equations
Sn = A1 - A1Rn
divided by 1 - R

R $$\neq 1$$

(I cant figured out how to make the "divided by" sign)

3. The attempt at a solution
Well, I subtracted 12 from 252 to find the sum of the last 4 terms and got 240. I could think of nothing more to do after this though.

2. Jun 11, 2010

### rock.freak667

If the sum of the first 6 number is 252 and the sum of the first two is 12. What does that say about the common ratio R? (more than 1 or less than 1?)

When you get that, then just apply your formula for Sn (Note that the formula you posted is valid for |R|<1)

3. Jun 11, 2010

### wvcaudill2

Okay, so the common ratio must be more than 1 since the numbers are getting bigger.

If my formula only works for |R|< 1, then how can I use that formula?

4. Jun 11, 2010

### rock.freak667

For |R|>1, the formula changes slightly

$$S_n = \frac{A(R^n-1)}{R-1}$$

Now S6 = 252 and S2= 12

5. Jun 11, 2010

### wvcaudill2

Okay, so do I just solve for R in one equation and then plug it into the other? I still have A though, which I assume I also need to find.

I am still unsure about how to go about solving this problem.

6. Jun 11, 2010

### rock.freak667

Well before you start solving, what are your two equations in A and R?

7. Jun 11, 2010

### rasmhop

This is incorrect. His formula is correct for all R not equal to 1. If you look at his formula and yours you will see that you just give him the exact same formula he posted himself (except you factored out A_1).

The condition |R|<1 is only required for infinite geometric series (i.e. when you let n go to infinity).

The rest of rock.freak667's advice is solid.

BTW the following manipulation may help you (the exact same information can be extracted from your formula):
$$S_2 = A+AR$$
$$S_6 = (A+AR) + (AR^2 + AR^3) + (AR^4 + AR^5) = (A+AR) + R^2(A+AR) + R^4(A+AR) = (1+R^2+R^4)(A+AR)$$

8. Jun 11, 2010

### wvcaudill2

Im not quite sure what you mean. But, when I reworked the first equation for R, I got:

R = 6th root of 252(R-1)+1 divided by A

and

R = square root of 12(R-1)+1 divided by A

I dont think this is the right direction though.

9. Jun 11, 2010

### rasmhop

You want to find R without refering to A so try dividing S_6 by S_2. You should get:
$$\frac{S_6}{S_2} = \frac{1-R^6}{1-R^2} = R^4 + R^2 + 1$$
which lets you solve for R^2.

10. Jun 11, 2010

### wvcaudill2

How did you get either of those equations for S6 or S2?

Earlier, you stated
S6 = (1 + R2 + R4)(A + AR)

and

S2 = (A + AR)

now, if you take S6/S2 would that not equal

(1 + R2 + R4)(A + AR)/(A + AR) = (1 + R2 + R4)

?

If so, then when solved for R I get R = the sixth root of 20, or about 1.6475. Is this correct?

11. Jun 11, 2010

### rasmhop

Factoring out the common factors from your formula we get:
$$S_n = \frac{A}{1-R}(1-R^n)$$
So in general:
$$\frac{S_n}{S_m} = \frac{ \frac{A}{1-R}(1-R^n)}{ \frac{A}{1-R}(1-R^m)} = \frac{1-R^n}{1-R^m}$$

Yes.

No. S_6/S_2 = 21 so you get:
$$R^4 + R^2 +1 = 21$$
or written differently:
$$R^4 + R^2 - 20 = 0$$
Which has the solution (the other solution is negative):
$$R^2 = \frac{-1+\sqrt{1+80}}{2} = \frac{-1+9}{2} = 4$$
Now you can find A from one of your equations and finally evaluate S_4.

12. Jun 11, 2010

### wvcaudill2

How did you get to R2 and its solution?

13. Jun 11, 2010

### rasmhop

If $x^2 + bx + c = 0$, then x is one of the following:
$$x_- = \frac{-b-\sqrt{b^2-4c}}{2} \qquad x_+ = \frac{-b+\sqrt{b^2-4c}}{2}$$
Have you not seen this? (I just assumed you knew it) If you haven't we may be able to find an alternative way to solve this.

14. Jun 11, 2010

### wvcaudill2

I thought I recognized the Quadratic formula, I just didnt think it could be used here.

I can solve for S4 now. Thank you for your help!