Question

if ax =by =cz, x is not equal to 0,y is not equal to 0and b2=ac then show that 1/x, 1/y,1/z are in ap​

Answer 1

Step-by-step explanation:

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)

hope this help you