You are given three positive numbers such that
i) A is the sum of the first two numbers.
ii) B is the sum of the first two numbers taken away from the third number.
iii) C is the sum of all these numbers.
iv)$$\dfrac{A}{B} = \dfrac{B}{C}$$
Select the correct option from below:

Let the 3 positive numbers be x, y, and z.

Given,

$$x + y = A$$
$$z - (x + y) = B$$
$$x + y + z = C$$

Also, $$\dfrac{A}{B}=\ \dfrac{B\ }{C}$$

$$\dfrac{x+y}{z-x-y}=\ \dfrac{z-x-y\ }{x+y+z}$$

Cross multiplying we get,

$$x^2+2xy+y^2+xz+yz\ =\ x^2+y^2+z^2-2xz+2xy-2yz$$

$$3x+3y=z$$

Thus, the sum of the first two numbers is equal to thrice the third number.

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You are given three positive numbers such thati) A is the sum of the first two numbers.ii) B is the sum of the first two numbers taken away from the third number.iii) C is the sum of all these numbers.iv)$$\dfrac{A}{B} = \dfrac{B}{C}$$Select the correct option from below: