# Mean and Third Proportional

We will learn how to find the mean and third proportional of the set of three numbers.

If x, y and z are in continued proportion then y is called the mean proportional (or geometric mean) of x and z.

If y is the mean proportional of x and z, y^2 = xz, i.e., y = +$$\sqrt{xz}$$.

For example, the mean proportion of 4 and 16 = +$$\sqrt{4 × 16}$$  = +$$\sqrt{64}$$ = 8

If x, y and z are in continued proportion then z is called the third proportional.

For example, the third proportional of 4, 8 is 16.

Solved examples on understanding mean and third proportional

1. Find the third proportional to 2.5 g and 3.5 g.

Solution:

Therefore, 2.5, 3.5 and x are in continuous proportion.

$$\frac{2.5}{3.5}$$ = $$\frac{3.5}{x}$$

⟹ 2.5x = 3.5 × 3.5

⟹ x = $$\frac{3.5 × 3.5}{2.5}$$

⟹ x = 4.9 g

2. Find the mean proportional of 3 and 27.

Solution:

The mean proportional of 3 and 27 = +$$\sqrt{3 × 27}$$ = +$$\sqrt{81}$$ = 9.

3. Find the mean between 6 and 0.54.

Solution:

The mean proportional of 6 and 0.54 = +$$\sqrt{6 × 0.54}$$ = +$$\sqrt{3.24}$$ = 1.8

4. If two extreme terms of three continued proportional numbers be pqr, $$\frac{pr}{q}$$; what is the mean proportional?

Solution:

Let the middle term be x

Therefore, $$\frac{pqr}{x}$$ = $$\frac{x}{\frac{pr}{q}}$$

⟹ x$$^{2}$$ = pqr × $$\frac{pr}{q}$$ = p$$^{2}$$r$$^{2}$$

⟹ x = $$\sqrt{p^{2}r^{2}}$$ = pr

Therefore, the mean proportional is pr.

5. Find the third proportional of 36 and 12.

Solution:

If x is the third proportional then 36, 12 and x are continued proportion.

Therefore, $$\frac{36}{12}$$ = $$\frac{12}{x}$$

⟹ 36x = 12 × 12

⟹ 36x = 144

⟹ x = $$\frac{144}{36}$$

⟹ x = 4.

6. Find the mean between 7$$\frac{1}{5}$$and 125.

Solution:

The mean proportional of 7$$\frac{1}{5}$$and 125 = +$$\sqrt{\frac{36}{5}\times 125} = +\sqrt{36\times 25}$$ = 30

7. If a ≠ b and the duplicate proportion of a + c and b + c is a : b then prove that the mean proportional of a and b is c.

Solution:

The duplicate proportional of (a + c) and (b + c) is (a + c)^2 : (b + c)^2.

Therefore, $$\frac{(a + c)^{2}}{(b + c)^{2}} = \frac{a}{b}$$

⟹ b(a + c)$$^{2}$$ = a(b + c)$$^{2}$$

⟹ b (a$$^{2}$$ + c$$^{2}$$ + 2ac) = a(b$$^{2}$$ + c$$^{2}$$ + 2bc)

⟹ b (a$$^{2}$$ + c$$^{2}$$) = a(b$$^{2}$$ + c$$^{2}$$)

⟹ ba$$^{2}$$ + bc$$^{2}$$ = ab$$^{2}$$ + ac$$^{2}$$

⟹ ba$$^{2}$$ - ab$$^{2}$$ = ac$$^{2}$$ - bc$$^{2}$$

⟹ ab(a - b) = c$$^{2}$$(a - b)

⟹ ab = c$$^{2}$$, [Since, a ≠ b, cancelling a - b]

Therefore, c is mean proportional of a and b.

8. Find the third proportional of 2x^2, 3xy

Solution:

Let the third proportional be k

Therefore, 2x^2, 3xy and k are in continued proportion

Therefore,

\frac{2x^{2}}{3xy} = \frac{3xy}{k}

⟹ 2x$$^{2}$$k = 9x$$^{2}$$y$$^{2}$$

⟹ 2k = 9y$$^{2}$$

⟹ k = $$\frac{9y^{2}}{2}$$

Therefore, the third proportional is $$\frac{9y^{2}}{2}$$.

● Ratio and proportion