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 \(\frac{9y^{2}}{2}\).












10th Grade Math

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