Suppose x-2, x and x+6 where are sequential terms in a geometric sequence

[Jaco Viljoen]

Homework Statement

Suppose x-2,x and x+6, where x is an integer, are consecutive terms in a geometric sequence S
Determine x

Homework Equations

r=x/(x-2)=(x+6)/x

The Attempt at a Solution

x/(x-2)=(x+6)/x cross multiply
x(x)=(x-2)(x+6)
x^2=x^2+4x-12
x^2-x^2+4x-12/4=0
4x=12
x=3

 

[SteamKing]

Homework Statement

Suppose x-2,x and x+6, where x is an integer, are consecutive terms in a geometric sequence S
Determine x

Homework Equations

r=x/x-2=x+6/x

The Attempt at a Solution

x/(x-2)=(x+6)/x cross multiply
x(x)=(x-2)(x+6)
x^2=x^2+4x-12
-4x+12/4=0
x=-3

You might want to double check your algebra when solving for x.

 

[Jaco Viljoen]

@SteamKing I have corrected the -to+, am I doing the right thing?
I am quite rusty with math, I have started a degree towards civil engineering now at the age of 33, I last practiced math in 2000.

Thank you for your input.
Jaco

 

[Ray Vickson]

Homework Statement

Suppose x-2,x and x+6, where x is an integer, are consecutive terms in a geometric sequence S
Determine x

Homework Equations

r=x/x-2=x+6/x

The Attempt at a Solution

x/(x-2)=(x+6)/x cross multiply
x(x)=(x-2)(x+6)
x^2=x^2+4x-12
x^2-x^2+4x-12/4=0
4x=12
x=3

Now is the time to learn to break bad habits: Never write something like x/x-2, because the means ##\frac{x}{x} - 2 = 1 - 2 = -1## when read using standard parsing rules for mathematical expressions. You want ##\frac{x}{x-2}##, so you need parentheses, like this: x/(x-2). Similarly for your other expression x+6/x, which means ##x + \frac{6}{x}## as you have written it. You should write (x+6)/x.

 

[Jaco Viljoen]

Thank you for pointing this out Ray,
I did pick up the error on the following statements before posting but have corrected these swell.
Does my answer make sense?

Thank you again.

 

[SammyS]

Does my answer make sense?

Thank you again.

What do you think?

Can you show that it does indeed make sense?

 

[HallsofIvy]

Apparently you have edited your original solution. You now have x= 3 in which case x- 2= 1, x= 3, and x+ 6= 9 so the sequence is 1, 3, 9 which is the same as [itex]3^0, 3^1, 3^2[/itex]. Yes, that is a geometric sequence. With you original answer, which was apparently x= -3, x- 2= -5, x= -3, x+ 6= 3. The sequence -5, -3, 3 is NOT a geometric sequence.

 

[SteamKing]

@SteamKing I have corrected the -to+, am I doing the right thing?
I am quite rusty with math, I have started a degree towards civil engineering now at the age of 33, I last practiced math in 2000.

Thank you for your input.
Jaco

Yes, your calculation is now correct.

A word of advice here. Always check your calculations. It's very easy for arithmetic mistakes to creep into a calculation. Making mistakes like this will be important to avoid later on in studying for your degree. Silly mistakes can cost you points on exams and assignments.

 

[Jaco Viljoen]

Thank you everyone,
I appreciate the help, distance learning has its downfall with math.
Have a great day.

 

[Jaco Viljoen]

a follow up question to this:
Suppose (x-2) is the 4th term, determine the first:
A_n=A₁*3^4-1
(x-2)=A₁*3³
(x-2)=A*27
(x-2)=27A
A=(x-2)/(27)
A=1/27 (x-2) was determined in the previous answer as 1
=3^-3