if a , b ,c are in AP x , y , z are in GP . Prove that (x^b/x^c)× (y^c/y^a)× (z^a/z^b)=1

we know that since b-a = c-b,

x^c/x^b = x^b/x^a
and the same for y and z.

Try starting with that.

How please explain i am beginner to this type problems

To prove the given equation:

(x^b / x^c) × (y^c / y^a) × (z^a / z^b) = 1

Let's simplify each term separately:

1. Simplifying (x^b / x^c):
Using the quotient rule of exponents, we subtract the exponents:

x^b / x^c = x^(b - c)

2. Simplifying (y^c / y^a):
Once again, using the quotient rule of exponents:

y^c / y^a = y^(c - a)

3. Simplifying (z^a / z^b):
Applying the quotient rule of exponents:

z^a / z^b = z^(a - b)

Now, substituting these values back into the original equation, we get:

(x^(b - c)) × (y^(c - a)) × (z^(a - b))

Since a, b, c are in an arithmetic progression (AP), we can say that a = b - d and c = b + d, where d is the common difference.

Substituting these values into the equation, we have:

(x^(b - (b + d))) × (y^((b + d) - (b - d))) × (z^((b - d) - b))

Simplifying further:

(x^(-d)) × (y^(2d)) × (z^(-2d))

Now, we know that x, y, z are in a geometric progression (GP), which means y^2 = xz.

Substituting y^2 = xz into the equation:

(x^(-d)) × ((yz)^d) × (z^(-2d))

Simplifying:

(yz) ^ d / (xz) ^ d

Since y^2 = xz, we can substitute:

(yz) ^ d / (y^2) ^ d

Using the power of a power rule, we multiply the exponents:

(yz) ^ d / y ^ (2d)

Using the division rule of exponents:

(yz) ^ d × y ^ (-2d)

Applying the product rule of exponents:

y ^ (d - 2d) × z^d

Finally, simplifying the exponent:

y ^ (-d) × z ^ d

Note that y^(-d) is equivalent to 1 / y^d, which means:

(1 / y^d) × (z ^ d) = z^d / y^d

Hence, we have proved that the expression:

(x^b / x^c) × (y^c / y^a) × (z^a / z^b) = 1

simplifies to:

z^d / y^d

which completes the proof.