Final answer:
To prove that 1/a + 1/c = 2/b, we substitute the given values of x, y, and z into the equation y² = xz and solve for x. By testing different values of x, we find that the equation is satisfied when x = y = z = 1 and b = 1.
Step-by-step explanation:
To prove that 1/a + 1/c = 2/b given x=y=zº and y² = xz, we can use the substitution method. Let's substitute x=y=zº in the equation y² = xz:
(zº)² = y(zº)
z² = zº
z² - zº = 0
z²(z - 1) = 0
z = 0 or z = 1
Since x=y=zº, we have x = 0 or x = 1
Now let's consider the case x = 0:
Since x = 0, y = 0 and z = 0
Substituting these values into 1/a + 1/c = 2/b:
1/0 + 1/0 = 2/b
This equation is undefined, so x = 0 doesn't satisfy the equation.
Now let's consider the case x = 1:
Since x = 1, y = 1 and z = 1
Substituting these values into 1/a + 1/c = 2/b:
1/1 + 1/1 = 2/b
1 + 1 = 2/b
2 = 2/b
b = 1
Therefore, the equation 1/a + 1/c = 2/b is satisfied when x = y = z = 1 and b = 1.