Question

ax=by=cz, if b2=ac then prove 2xz/x+z

Answer 1

Given ax = by = cz and b2 = ac
Let ax = by = cz = k
Consider, ax = k ⇒ a = k1/x
Similarly, we get b = k1/y and c = k1/z
Now consider, b2 = ac
⇒ (k1/y)2 = (k1/x)(k1/z)
⇒ k2/y = k1/x + 1/z
Comparing both the sides, we get
(2/y) = (1/x) + (1/z)
⇒ (2/y) = (x + z)/xz
∴ y = (2xz)/(x + z)

Answer 2

Answer:

Given ax = by = cz and b2 = ac

Let ax = by = cz = k

Consider, ax = k ⇒ a = k1/x

Similarly, we get b = k1/y and c = k1/z

Now consider, b2 = ac

⇒ (k1/y)2 = (k1/x)(k1/z)

⇒ k2/y = k1/x + 1/z

Comparing both the sides, we get

(2/y) = (1/x) + (1/z)

⇒ (2/y) = (x + z)/xz

∴ y = (2xz)/(x + z)

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