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Question
If ax = by = cz and b2 = ac, prove that: y = `[2xz]/[x + z]`
Solution
Let ax = by = cz = k
∴ a = `k^(1/x) ; b = k^(1/y) ; c = k^(1/z)`
Also, We have b2 = ac
∴ `( k^(1/y))^2 = ( k^(1/x)) xx ( k^(1/z))`
⇒ `k^(2/y) = k^( 1/x + 1/z )`
⇒ `k^(2/y) = k^[ z + x ]/[ xz ]`
Comparing the powers we have
`2/y = [ z + x ]/[ xz ]`
⇒ `y = [ 2 xz ]/[ z + x ]`
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