Question

Find two numbers whose mean proportion is 36 and the third proportional is 288.

Answer 1

Let the two numbers be $$a$$ and $$b$$.

Mean proportion is 36:

$$\sqrt{ab} = 36$$

$$\Rightarrow ab = 36^2 = 1296$$ 

$$\Rightarrow a = \frac{1296}{b} \quad \text{... (1)}$$

Third proportional is 288:

$$a : b :: b : 288$$

$$\Rightarrow b^2 = 288a \quad \text{... (2)}$$

Substitute (1) into (2):

$$b^2 = 288 \times \frac{1296}{b}$$

$$b^3 = 288 \times 1296$$

$$= (2 \times 144) \times (9 \times 144)$$

$$= 18 \times 144^2$$

$$= 373248$$

Since $$288 = 2 \times 12^2$$ and $$1296 = 12^2 \times 9$$:

$$b^3 = 12^3 \times (2 \times 9 \times 12)$$

$$= 12^3 \times 216$$

$$= 12^3 \times 6^3$$

$$= (72)^3$$

$$b = 72$$

Finding $$a$$:

$$a = \frac{1296}{72} = 18$$

Hence, the two numbers are 18 and 72.