Find two numbers whose mean proportional is 15 and the third proportional is 405

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

10th Grade Math

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Find two numbers whose mean proportional is 16 and the third proportional is 128.

Let x and y be two numbersTheir mean proportion = 16and third proportion = 128∴ `sqrt(xy)` = 16⇒ xy = 256⇒ x = `(256)/y`      ....(i)and `y^2/x` = 128⇒ x = `y^2/(128)`     ....(ii)From (i) and (ii)`(256)/y = y^2/(128)`

⇒ y3 = 256 x 128

⇒ y3 = (32)3

∴ Numbers are 8, 32.

Concept: Concept of Proportion

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Find two numbers such that the mean proportional between them is 12 and the third proportional to them is 96.

Let a and b be the two numbers whose mean proportional is 12

`∴ ab = 12^2 => ab = 144 => 144/a` ....(1)

Now third proportional is 96

∴ a : b : : b : 96

`=> b^2 = 96a`

`=> (144/a)^2 = 96a`

`=> a^3 = (144 xx 144)/96`

`=> a^3 = 216` 

`=> a = 6`

`b = 144/6 = 24`

Therefore the number are 6 and 24.

Concept: Concept of Proportion

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