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.
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
==> 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.