# Geometric Sequences & Series

wvcaudill2

## Homework Statement

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.

## Homework Equations

Sn = A1 - A1Rn
divided by 1 - R

R $$\neq 1$$

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

## 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.

## Answers and Replies

Homework Helper
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)

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?

Homework Helper
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?

For |R|>1, the formula changes slightly

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

Now S6 = 252 and S2= 12

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.

Homework Helper
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.

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

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

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):
$$S_2 = A+AR$$
$$S_6 = (A+AR) + (AR^2 + AR^3) + (AR^4 + AR^5) = (A+AR) + R^2(A+AR) + R^4(A+AR) = (1+R^2+R^4)(A+AR)$$

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.

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.

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?

rasmhop
How did you get either of those equations for S6 or S2?
Factoring out the common factors from your formula we get:
$$S_n = \frac{A}{1-R}(1-R^n)$$
So in general:
$$\frac{S_n}{S_m} = \frac{ \frac{A}{1-R}(1-R^n)}{ \frac{A}{1-R}(1-R^m)} = \frac{1-R^n}{1-R^m}$$

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)

?
Yes.

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

wvcaudill2
How did you get to R2 and its solution?

rasmhop
How did you get to R2 and its solution?

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.

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!