If \(a, b, c\) are in G.P. and \(a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}}\), then which one of the following is correct?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2y=x+z\)
Three numbers are in G.P. when the square of the middle one equals the product of the other two: \(b^2=ac\). Writing \(a\), \(b\), \(c\) as powers of one common base turns this into a relation between the powers.
Let each of the equal quantities be \(k\): \(a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}=k\)
Then \(a=k^{x}\), \(b=k^{y}\), \(c=k^{z}\)
\(b^2=ac\) gives \(k^{2y}=k^{x}\times k^{z}=k^{x+z}\)
Equal powers of the same base: \(2y=x+z\)
So \(x\), \(y\), \(z\) are in A.P.