Question

if a^1/x=b^1/y=c^1/z and b^2=ac then x+z=​

Answer 1

ANSWER:

Given:

• a^(1/x) = b^(1/y) = c^(1/z)
• b² = ac

To Find:

• x + z

Solution:

We are given that,

$:\implies a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}$

Let these be equal to a constant k, i.e.,

$:\implies a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}=k$

So,

$:\implies a^{\frac{1}{x}}=k\:\:\:\implies a=k^x- - - -(1)$

Similarly,

$:\implies b^{\frac{1}{y}}=k\:\:\:\implies b=k^y- - - -(2)$

And,

$:\implies c^{\frac{1}{z}}=k\:\:\:\implies c=k^z- - - -(3)$

Now, we are also given that,

$:\implies b^2=ac$

Putting value of a, b and c from (1), (2) & (3),

$:\implies (k^y)^2=(k^x)(k^z)$

So,

$:\implies k^{2y}=k^{x+z}$

Now, comparing the powers because the bases are same,

$:\implies 2y=x+z$

Hence,

$:\implies\bf x+z=2y$