1. ## Geometric Sequence Problem

"The sum of the first 2 terms of a geometric sequence is four times the sum of the next 2 terms, and the sum to infinity is 3. Determine 'r' and write down the sequences."

I don't know how to start. Any pointers for what I need to do? Solutions welcome, this is prep for a test I'm writing in two hours. Thanks in advance.

2. ## Re: Geometric Sequence Problem

Some equations to consider $a+ar = 4(ar^2+ar^3)$ and $\frac{a}{1-r}= 3$

3. ## Re: Geometric Sequence Problem

a + ar = 4(ar^2 + ar^3)
a(1 + r) = 4ar^2(1 + r)
a = 4ar^2
1 = 4r^2
r^2 = 1/4
r = 1/2

4. ## Re: Geometric Sequence Problem

Hello, Pulsys!

The sum of the first 2 terms of a geometric sequence is four times the sum of the next 2 terms,
and the sum to infinity is 3. .Determine $r$ and write down the sequences.

The first four terms are: . $a,\,ar,\,ar^2,\,ar^3$

We are told: . $\begin{Bmatrix}a + ar \:=\:4(ar^2 + ar^3)& [1] \\ \dfrac{a}{1-r} \:=\:3 & [2] \end{Bmatrix}\;\;\text{ Note that: }\,|r| < 1$

From [2], we have: . $a \:=\:3(1-r)\;\;[3]$

From [1], we have: . $a+ar \:=\:4ar^2 + 4ar^3 \quad\Rightarrow\quad 4ar^3 + 4ar^2 - ar - a \:=\:0$

. . . $4ar^2(r+1) - a(r+1) \:=\:0 \quad\Rightarrow\quad a(r+1)(4r^2 - 1) \:=\:0$

. . $a(r+ 1)(2r-1)(2r+1) \:=\:0 \quad\Rightarrow\quad r \:=\:\rlap{//}\text{-}1,\:\tfrac{1}{2},\:\text{-}\tfrac{1}{2}$

Substitute into [3]: . $\begin{Bmatrix}r = \tfrac{1}{2}\!: & a = \tfrac{9}{2} \\ \\[-3mm] r = \text{-}\tfrac{1}{2}\!: & a = \tfrac{3}{2} \end{Bmatrix}$

The sequences are: . $\begin{Bmatrix} \frac{9}{2},\:\frac{9}{4},\:\frac{9}{8},\:\frac{9} {16}\:\hdots \\ \\[-3mm] \frac{3}{2},\:\text{-}\frac{3}{4},\:\frac{3}{8},\:\text{-}\frac{3}{16}\:\hdots \end{Bmatrix}$