Four numbers are in proportion. The sum of the squares of the four numbers is 50 and the sum of the mean is 5. The ratio of first two terms is 1 : 3. What is the average of the four numbers?

Let the four numbers be 'a', 'b', 'c' and 'd'

Given ratio of first two numbers is 1:3 and four numbers are in proportion,

$$\therefore$$ a : b = c : d = 1 : 3 

$$\Rightarrow$$ b = 3a and d = 3c

Sum of the squares of the four numbers = 50

$$\Rightarrow$$ a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ + d$$^{2}$$ = 50

$$\Rightarrow$$ a$$^{2}$$ + 9a$$^{2}$$ + c$$^{2}$$ + 9c$$^{2}$$ = 0

$$\Rightarrow$$ 10a$$^{2}$$ + 10c$$^{2}$$ = 0

$$\Rightarrow$$ a$$^{2}$$ + c$$^{2}$$ = 0........(1)

Sum of the means = 5 

b + c = 5 

3a + c = 5 (as b = 3a)

a = $$\frac{5 - c}{3}$$........................(2)

From equations (1) and (2)

$$(\frac{5 - c}{3})^{2} + c^{2} = 0$$ 

$$\frac{25 + c^{2} - 10c}{9} + c^{2} = 5$$

$$25 + c^{2} - 10c + 9c^{2} = 5$$

$$10c^{2} - 10c + 20 = 0$$

$$c^{2} - c + 10 = 0$$

After solving the equation we get, c = 2.

Substitute the value of 'c' in equation (2)

a = $$\frac{5 - 2}{3}$$ (or) a = 1

We know b = 3a (or) b = 3

We know d = 3c (or) d = 6

Average of the four numbers,

= $$\frac{1 + 3 + 2 + 6}{4}$$

= 3

Hence, option B is the correct answer.