find x if x-3, 3x+5 and 18x-5 are three consecutive term of a geometric progression
5 answers
how can we find x in x-3, 3x+5 and 18x-5 geometric progression
(x - 3) r =(3 x + 5)
(3 x + 5) r = 18 x - 5
(3 x + 5)/(x - 3) = ( 18 x - 5) / (3 x + 5 )
(3 x + 5)^2 = (x - 3)( 18 x - 5)
9 x^2 + 30 x + 25 = 18 x^2 - 59 x + 15
9 x^2 - 89 x - 10 = 0
(x-10)(9x+1) = 0
x = 10 or x = -1/9
using x = 10
r = 35/7 = 5
or
r = 175/35 = 5
that works
(3 x + 5) r = 18 x - 5
(3 x + 5)/(x - 3) = ( 18 x - 5) / (3 x + 5 )
(3 x + 5)^2 = (x - 3)( 18 x - 5)
9 x^2 + 30 x + 25 = 18 x^2 - 59 x + 15
9 x^2 - 89 x - 10 = 0
(x-10)(9x+1) = 0
x = 10 or x = -1/9
using x = 10
r = 35/7 = 5
or
r = 175/35 = 5
that works
Remember that there is a common ratio. That means that
(3x+5)/(x-3) = (18x-5)/(3x+5)
Now just solve for x.
(3x+5)/(x-3) = (18x-5)/(3x+5)
Now just solve for x.
use the properties of a GP
(3x+5)/(x-3) = (18x-5)/(3x+5)
(x-3)(18x-5) = (3x+5)^2
18x^2 - 69x + 15 = 9x^2 + 30x + 25
9x^2 - 89x - 10 = 0
(x-10)(9x + 1) = 0
x = 10 or x = -1/9
(3x+5)/(x-3) = (18x-5)/(3x+5)
(x-3)(18x-5) = (3x+5)^2
18x^2 - 69x + 15 = 9x^2 + 30x + 25
9x^2 - 89x - 10 = 0
(x-10)(9x + 1) = 0
x = 10 or x = -1/9
It is okay
11 answers