# [SOLVED] geometric series question

• Apr 5th 2009, 03:02 PM
brentwoodbc
[SOLVED] geometric series question
if x-1, x+3 and x+12 are consecutive terms in a geometric sequence what is the ratio.

I was told you put t3 over t2, and also t2 over t1 and cross multiply, which gives you a quadratic, but when I do it my (a) value is 0 so it would be

5 +/-5 over 0
?

then you sub that into the original equation? how do you sub x into tn=ar^n-1
or sn=a(r^n-1over r-1)?

thanks.
• Apr 5th 2009, 03:09 PM
stapel
Quote:

Originally Posted by brentwoodbc
if x-1, x+3 and x+12 are consecutive terms in a geometric sequence what is the ratio.

I was told you put t3 over t2, and also t2 over t1 and cross multiply, which gives you a quadratic, but when I do it my (a) value is 0 so it would be

5 +/-5 over 0

How? What were your steps? (We can't find the errors until you show your work.) (Wink)

By nature of geometric sequences, yes, you need to set up the ratios. And since the ratio is "common", you get:

. . . . .$\displaystyle \frac{x\, +\, 12}{x\, +\, 3}\, =\, \frac{x\, +\, 3}{x\, -\, 1}$

Then you need to solve the rational equation.

. . . . .$\displaystyle (x\, -\, 1)(x\, +\, 12)\, =\, (x\, +\, 3)(x\, +\, 3)$

. . . . .$\displaystyle x^2\, +\, 11x\, -\, 12\, =\, x^2\, +\, 6x\, +\, 9$

. . . . .$\displaystyle 11x\, -\, 12\, =\, 6x\, +\, 9$

I'm not seeing how you arrived at division by zero...? Please reply showing your steps in solving the linear equation.

Thank you! :D
• Apr 5th 2009, 05:25 PM
brentwoodbc
Quote:

Originally Posted by stapel
How? What were your steps? (We can't find the errors until you show your work.) (Wink)

By nature of geometric sequences, yes, you need to set up the ratios. And since the ratio is "common", you get:

. . . . .$\displaystyle \frac{x\, +\, 12}{x\, +\, 3}\, =\, \frac{x\, +\, 3}{x\, -\, 1}$

Then you need to solve the rational equation.

. . . . .$\displaystyle (x\, -\, 1)(x\, +\, 12)\, =\, (x\, +\, 3)(x\, +\, 3)$

. . . . .$\displaystyle x^2\, +\, 11x\, -\, 12\, =\, x^2\, +\, 6x\, +\, 9$

. . . . .$\displaystyle 11x\, -\, 12\, =\, 6x\, +\, 9$

I'm not seeing how you arrived at division by zero...? Please reply showing your steps in solving the linear equation.

Thank you! :D

cross multiply and get $\displaystyle x^2+6x+p=x^2+11x-12$
which is
$\displaystyle 0x^2-5x+21$

5 +/- sqrt5 over 2x0

=
5 {+/-} 5 over 0

I dont know why my teacher said to use the quadratic equation because you can just solve for x, but once you have x how do you find r?
• Apr 5th 2009, 05:30 PM
brentwoodbc
Quote:

Originally Posted by stapel
How? What were your steps? (We can't find the errors until you show your work.) (Wink)

By nature of geometric sequences, yes, you need to set up the ratios. And since the ratio is "common", you get:

. . . . .$\displaystyle \frac{x\, +\, 12}{x\, +\, 3}\, =\, \frac{x\, +\, 3}{x\, -\, 1}$

Then you need to solve the rational equation.

. . . . .$\displaystyle (x\, -\, 1)(x\, +\, 12)\, =\, (x\, +\, 3)(x\, +\, 3)$

. . . . .$\displaystyle x^2\, +\, 11x\, -\, 12\, =\, x^2\, +\, 6x\, +\, 9$

. . . . .$\displaystyle 11x\, -\, 12\, =\, 6x\, +\, 9$

I'm not seeing how you arrived at division by zero...? Please reply showing your steps in solving the linear equation.

Thank you! :D

I dont know why he said to use the quadratic formula because you dont need it, and it doesnt work.

so I sub the 4.2 into the x and I get the answer of 2.25.