if (3-x)=6, (7-5x) are consecutive terms of GP with common ratio r>0, fing d value of (x) and d common ratio.

thanks

R = u2/u1 = u2/u2 = u4/u3

6/3-x = 7-5x/6

You will need brackets in

6/(3-x) = (7-5x)/6
and
(3-x)(7-5x) = 36
21 -22x + 5x^2 = 36
5x^2 - 22x - 15 = 0
(x - 5)(5x + 3) = 0
x = 5 or x = -3/5

the common ratio is 6/(3-x)

if x =5, common ratio is 6/-2 or -3
if x = -3/5, common ratio is 6/(3 + 3/5) or 5/3

Fayokemi

To find the values of x and the common ratio, we can solve the given equations.

First, let's simplify the equations:

(3 - x) = 6 --(1)
(7 - 5x) = r * (3 - x) --(2)

From equation (1), we can find the value of x.

3 - x = 6
Subtract 3 from both sides:
-x = 3

Multiply both sides by -1 to isolate x:
x = -3

Now, substitute the value of x into equation (2) to find the common ratio, r:

(7 - 5x) = r * (3 - x)

(7 - 5(-3)) = r * (3 - (-3))
(7 + 15) = r * (3 + 3)
22 = r * 6

Divide both sides by 6 to solve for r:
r = 22/6
r = 11/3

Therefore, the value of x is -3, and the common ratio is 11/3.