If x-6,x-2 and x+10 are three consecutive terms of a geometric progression, find the value of x​

Step-by-step explanation:

let the common ratio be r

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

(x-2)r=(x+10)

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

and r = (x+10)/(x-2)

so, (x-2)/(x-6)=(x+10)/(x-2)

sloving it, you'll get x=8

so the value of x is 8

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0