Question

If 1/x,1/y,1/z are in ap,show that xy,zx,yz are in ap​

Answer 1

Answer:

*Proof below

Step-by-step explanation:

Dude its very easy....just follow my steps...

we know, 2b = a + c [if a, b, c are in A.P.]....(i)

now since, the terms in your question are in A.P.,

$= ) 2 \times \frac{1}{y} = \frac{1}{x} + \frac{1}{z}$

$= ) \frac{2}{y} = \frac{z + x}{xz}$

$= )2xz = y(z + x)$

$= )2xz = zy + xy$

Hence from the same reasoning of (i),

xy, zx, yz are in A.P. (proved)

Hope it helps you buddy....

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