# If x,y,z are the terms of an arithmetic sequence prove that 1/y*z , 1/z*x, 1/x*y are in a.p.

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x,y, z are terms in an arthimnatic progression,

We need to prove that :

1/y*z , 1/ z*x , 1/x*y are in the progresion

In the above terms are in the progresion, then:

1/z*x - 1/y*z = 1/x*y - 1/z*x.

We know that:

y- x = z - y

Let us divide by x*y*z

==> (y-x)/x*y*z = (z-y)/x*y*z

==> y/xyz - x/ xyz = z/xyz - y/ xyz

Reduce similar:

==> 1/xz - 1/yz = 1/xy - 1/xz

Then the terms belong to the progression.

Since x,y and z are in arithmetic progression,

So the difference of the successive terms should be equal

y-x = z-y..........(1)

So the terms 1/yz , 1/zx and 1/xy are equal to x/xyz, y/xyz and z/xyz. respectively.

So the dfference of suceesive terms is

(y-x)/xyz and (z-x)/xyz which are equal as the merator are equal by (1).

So 1/yz , 1/zx and 1/xy are in AP.

If x,y,z are the consecutive terms of an A.P., then;

y = (x+z)/2

If we have to prove that 1/y*z , 1/z*x, 1/x*y are also the terms of an A.P., we'll have to demonstrate that:

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

To add the ratios from the numerator, we'll have to calculate the LCD of them:

LCD = x*y*z

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

So the relation that has to be demonstrated will become:

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

But, (x+z)/2 = y, so, we'll substitute the relation into the calculated expression:

1/z*x = y/x*y*z

We'll reduce like terms:

1/z*x = 1/z*x

**So, the terms 1/y*z , 1/z*x, 1/x*y are in A.P.**