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