if a^x=b^y=c^z and b2=ac show that 1/x , 1/y , 1/z are in ap
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Answered by
6
Answer:
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)
Answered by
2
Answer:
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)
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