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.