Question

If a^1/x = b^1/y = c^1/z and b^2=ac and prove that x+ y +z=3y

Answer 1

Answer:

Let a^1/x=b^1/y=c^1/z= k

So, a=k^x,b=k^y and c= k^z

Then using b²=ac

k^2y=k^x+z

2y=x+z

Therefore,x+y+z=3y. [Hence proved]

I hope it helps you.☺

Step-by-step explanation:

Answer 2

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Step-by-step explanation:

Let us consider that

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

Then, we have

$a^\frac{1}{x}=k~~~~~\Rightarrow a=k^x,\\\\b^\frac{1}{y}=k~~~~~\Rightarrow b=k^y\\\\c^\frac{1}{z}=k~~~~~\Rightarrow c=k^z$

Also, it is given that

$b^2=ac~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We know the following properties of exponents :

$(a)~p^r.p^s=p^{r+s},\\\\(b)~p^r=p^s~~~\Rightarrow r=s.$

Therefore, from equation (i), we get

$(k^y)^2=k^x.k^z\\\\\Rightarrow k^{2y}=k^{x+z}\\\\\Rightarrow 2y=x+z\\\\\Rightarrow 2y+y=x+z+y\\\\\Rightarrow x+y+z=3y.$

Hence proved.