Question

If a^x=b^y=c^z a,b,c not equal to 0.b^2=ac.then prove that 1/x,1/y,1/z are in AP

Answer 1

Note:

★ AP ( Arithmetic Progression ) : A sequence is said to be in AP if the difference between the consecutive terms are equal .

★ If a , b , c are in AP , then 2b = a + c

★ The general term of an AP is given by ;

T(n) = a + (n - 1)d , where a is the first term and d is the common difference .

★ The general form of an AP is given by ;

a , (a + d) , (a + 2d) , (a + 3d) , ... , [a + (n - 1)d]

Solution:

Given : a^x = b^y = c^z , b² = ac

To prove : 1/x , 1/y , 1/z are in AP

ie ; 2/y = 1/x + 1/z

Proof :

Let a^x = b^y = c^z = k

If a^x = k , then

a = k^(1/x)

If b^y = k , then

b = k^(1/y)

If c^z = k , then

c = k^(1/z)

Also,

=> b² = ac

=> [ k^(1/y) ]² = [ k^(1/x) ]•[ k^(1/y) ]

=> k^(2/y) = k^(1/x + 1/y)

=> 2/y = 1/x + 1/y

=> 1/x , 1/y , 1/z are in AP

Thus,

1/x , 1/y , 1/z are in AP .

Hence proved .