Geometric Sequences and Series

[Guest]

id like some help with these questions please

1) If x-3, x-1, and 4x-2 are consecutive terms in a geometric sequence, find x

Answers are 1/3 and 5, I can solve for 1/3 but not 5, I must be solving the wrong way

2) The sum of the first 2 terms of a geometric sequence is 3. The sum of the next 2 terms is 4/3. Find the first 4 terms

Answers are 9/5, 6/5, 4/5, 8/15 or 9. -6, 4, -8/3

these are giving me trouble, thanks

 

[stapel]

It will be a lot easier to check your work once you've posted it.

Please be specific. Thank you.

Eliz.

 

[soroban]

Hello, a456345!

Here's the second one . . .

The sum of the first two terms of a geometric sequence is \(\displaystyle 3\).
The sum of the next two terms is \(\displaystyle \frac{4}{3}\).
Find the first four terms.

Answers: \(\displaystyle \L\frac{9}{5},\:\frac{6}{5},\:\frac{4}{5},\:\frac{8}{15}\,\) or \(\displaystyle \L\,9,\,-6,\,4,\,-\frac{8}{3}\)

The first four terms of a geometric series are: \(\displaystyle \:a,\:ar,\:ar^2,\:ar^3\)

So we have: \(\displaystyle \:\begin{array}{cccccccc}a\,+\,ar & = & 3 & \;\;\Rightarrow\;\; & a(1\,+\,r) & = & 3 & \;(1) \\
ar^2\,+\,ar^3 & = & \frac{4}{3} & \:\Rightarrow\: & ar^2(1\,+\,r) & = & \frac{4}{3} & \2)\end{array}\)

Divide (2) by (1): \(\displaystyle \L\:\frac{ar^2(1\,+\,r)}{a(1\,+\,r)} \:=\:\frac{\frac{4}{3}}{3}\)\(\displaystyle \;\;\Rightarrow\;\;r^2\,=\,\frac{4}{9}\;\;\Rightarrow\;\;r\:=\:\pm\frac{2}{3}\)

If \(\displaystyle r\,=\,\frac{2}{3}\), then: \(\displaystyle a\,=\,\frac{9}{5}\)
. . The series begins: \(\displaystyle \L\:\frac{9}{5},\:\frac{6}{5},\:\frac{4}{5},\:\frac{8}{15}\)

If \(\displaystyle r\,=\,-\frac{2}{3}\), then: \(\displaystyle a\,=\,9\)
. . The series begins: \(\displaystyle \L\:9,\:-6,\:4,\:-\frac{8}{3}\)

 

[Guest]

id like some help with these questions please

1) If x-3, x-1, and 4x-2 are consecutive terms in a geometric sequence, find x

Answers are 1/3 and 5, I can solve for 1/3 but not 5, I must be solving the wrong way

is this right?

(x+1)/(x-3) = (4x-2)/(x+1)

 

[Guest]

I figured the above question out
You cross multiply what I had earlier until you end up with a quadratic eqn, then find the x intercepts

need help with these

The sum of the first two terms of a geometric series is 12 and the sum of the first three terms is 62. Find the first three terms

Answers are 2, 10, 50 and 72, -60, 50

In the geometric series, a = 3 and S3 = 21, find the common ratio and sum of first 7 terms

answers are ratio is 2 and sum is 381

thanks i really appreciate it

 

[Denis]

>The sum of the first two terms of a geometric series is 12 and the sum of
>the first three terms is 62. Find the first three terms
>Answers are 2, 10, 50 and 72, -60, 50

Hint: 62 - 12 = 50 = 3rd term

>In the geometric series, a = 3 and S3 = 21, find the common ratio and sum of
>first 7 terms
>answers are ratio is 2 and sum is 381

Hint: 3 + 3x + 3x^2 = 21
(2 solutions)

 

[Guest]

Thank you for your help I finished the 2nd question

Iam still not seeing this though

The sum of the first two terms of a geometric series is 12 and the sum of
the first three terms is 62. Find the first three terms
Answers are 2, 10, 50 and 72, -60, 50

Hint: 62 - 12 = 50 = 3rd term

Can someone show me how to find a and r?

 

[soroban]

Hello, a456345!

The sum of the first two terms of a geometric series is 12
and the sum of the first three terms is 62. .Find the first three terms

Answers: \(\displaystyle \{2,\,10,\,50\}\,\) and \(\displaystyle \,\{72,\,-60,\,50\}\)

The first three terms are: \(\displaystyle \:a,\:ar,\:ar^2\)

We have: \(\displaystyle \:\begin{array}{cccccccc}a\,+\,ar & \,=\, & 12 & \;\Rightarrow\; & a(1\,+\,r) & \;=\; & 12 & \;(1)\\
a\,+\,ar\,+\,ar^2 & \,=\, & 62 & \Rightarrow & a(1\,+\,r\,+\,r^2) & = & 62 & \;(2)
\end{array}\)

Divide (2) by (1): \(\displaystyle \L\:\frac{a(1\,+\,r\,+\,r^2)}{a(1\,+\,r)} \:=\:\frac{62}{12}\;\;\Rightarrow\;\;\frac{1\,+\,r\,+\,r^2}{1\,+\,r}\:=\:\frac{31}{6}\)

We have: \(\displaystyle \:6\,+\,6t\,+\,6r^2\:=\:31\,+\,31r\;\;\Rightarrow\;\;6r^2\,-\,25r\,-\,25\:=\:0\)

. . which factors: \(\displaystyle \r\,-\,5)(6r\,+\,5)\:=\:0\)

. . and has roots: \(\displaystyle \:r\:=\:5,\,-\frac{5}{6}\)

. . . . . . . . . then: \(\displaystyle \:a\:=\:2,\:72\)


There are two possible sequences: \(\displaystyle \:2,\,10,\,50\,\) and \(\displaystyle \,72,\,-60,\,50\)

In a geometric series: \(\displaystyle \,a\,=\,3,\;S_3\,=\,21\)
Find the common ratio and sum of first 7 terms.

Answers: \(\displaystyle \,r \,=\,2,\:S_7\,=\,381\)

I found two answers for this one . . .


We are given: \(\displaystyle \:a\,=\,3\)

The first three terms are: \(\displaystyle \:3,\:3r,\:3r^2\)

Their sum is: \(\displaystyle \:3\,+\,3r\,+\,3r^2\:=\:21\;\;\Rightarrow\;\;1\,+\,r\,+\,r^2\:=\:7\)

We have: \(\displaystyle \:r^2 \,+\,r\,-\,6\:=\:0\;\;\Rightarrow\;\;(r\,-\,2)(r\,+\,3)\:=\:0\;\;\Rightarrow\;\;r\:=\:2,\,-3\)


With \(\displaystyle a\,=\,3,\:r\,=\,2\), we have: \(\displaystyle \:S_7\:=\:381\)

With \(\displaystyle a\,=\,3,\:r\,=\,-3\), we have: \(\displaystyle \:S_7\:=\:1641\)