Determine two geometric sequences whose first three terms are 18x-9, 2x+8, and x-1.
2006-10-01 14:36:33 · 4 answers · asked by gg 4 in Science & Mathematics ➔ Mathematics
This is not solving for X,
2006-10-01 14:48:49 · update #1
If these three terms form a geometric series, then some number (call it "y") times the first number is the second number and that same number times the second number is the third number
(18x-9) y = 2x+8
(2x+8) y = x-1
So now you have two equations in two variables. From the first one you get...
y = (2x+8)/(18x-9)
plug that into the second equation...
(2x+8)^2 / (18x-9) = x-1
4x^2 + 32x + 64 = 18x^2 - 27x + 9
0 = 14x^2 - 59x - 55
Solve using the quadratic formula or factoring into ...
0 = (x-5)(14x+11)
x = 5 or -11/14
When x=5, the numbers are 81, 18, 4
I'll let you calculate the values for -11/14
2006-10-01 14:57:52 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Do you start the series with n=0 or n =1 (check calc book for notation)? (Assuming the notation SUM of a*r^n)
If you start with n=0, then a=18x-9, ar=2x+8, ar^2=x-1. So you need to find r. r=(2x+8)/(18x-9).
If you start with n=1, then ar=18x-9, ar^2=2x+8, ar^3=x-1. So try to find r as before: ar^2=2x+8 or =ar*r=(18x-9)*r so divide (2x+8)/(18x-9) to get r.
2006-10-01 21:59:23 · answer #2 · answered by raz 5 · 0⤊ 0⤋
do you want to find out what x is or something?
well if you do, set them equal to each other
18x - 9 = x - 1
18x - 9 = 2x + 8
2x + 8 = x - 1
and solve for x, simple
2006-10-01 21:40:58 · answer #3 · answered by sur2124 4 · 0⤊ 1⤋
boring too easy
2006-10-01 21:43:46 · answer #4 · answered by porro 2 · 0⤊ 2⤋