Question

Please solve the below one:

Answer 1

HELLO DEAR,


GIVEN THAT:-

a^x = b^y = c^z = k(say)

$\mathbf{a = k^{1/x} , b = k^{1/y} and c = k^{1/z}}$

b² = ac

$\mathbf{K^{2/y} = k^{1/x}*k^{1/z}}$

$\mathbf{K^{2/y} = k^{(x + z)/xz}}$

2/y = (x + z)/xz

2xz = xy + yz-------( 1 )

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

WE KNOW THAT:-

common difference = $\mathbf {a_2 - a_1 = a_3 - a_2}$

1/y - 1/x = 1/z - 1/y

(X - y)/xy = (y - z)/zy

Zy(x - y) = xy(y - z)

Xyz - zy² = xy² - xyz

2xyz = y(xy + yz)

2xz = (xy + yz)
On comparing with ------( 1 )

Hence,

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

I HOPE ITS HELP YOU DEAR,
THANKS