Problem Analysis

We are given three numbers: 16, 20, and 30. A specific number is added to each, making the new numbers form a continued proportion. Our goal is to find this number, identify the largest and smallest of the resulting numbers, and then calculate their mean proportional.

Finding the Added Number ($x$)

Let the number added to each term be $x$. The resulting sequence is $16+x$, $20+x$, and $30+x$. For these numbers to be in continued proportion, the ratio between the first two must equal the ratio between the second two:

$ \frac{16+x}{20+x} = \frac{20+x}{30+x} $

To solve for $x$, we cross-multiply:

$ (16+x)(30+x) = (20+x)^2 $

Expand both sides of the equation:

$ 16 \times 30 + 16x + 30x + x^2 = 20^2 + 2(20)x + x^2 $

$ 480 + 46x + x^2 = 400 + 40x + x^2 $

Simplify by cancelling $x^2$ from both sides and rearranging terms to solve for $x$:

$ 480 + 46x = 400 + 40x $

$ 46x - 40x = 400 - 480 $

$ 6x = -80 $

$ x = \frac{-80}{6} = \frac{-40}{3} $

Determining Resulting Numbers

Now, substitute $x = \frac{-40}{3}$ back into the expressions for the resulting numbers:

• First number: $16 + (\frac{-40}{3}) = \frac{48}{3} - \frac{40}{3} = \frac{8}{3}$
• Second number: $20 + (\frac{-40}{3}) = \frac{60}{3} - \frac{40}{3} = \frac{20}{3}$
• Third number: $30 + (\frac{-40}{3}) = \frac{90}{3} - \frac{40}{3} = \frac{50}{3}$

The resulting numbers are $\frac{8}{3}$, $\frac{20}{3}$, and $\frac{50}{3}$. The smallest is $\frac{8}{3}$ and the largest is $\frac{50}{3}$.

Calculating the Mean Proportional

The mean proportional ($M$) between two numbers $a$ and $b$ is calculated as $M = \sqrt{a \times b}$. We need the mean proportional between the smallest ($\frac{8}{3}$) and the largest ($\frac{50}{3}$) resulting numbers.

$ M = \sqrt{\frac{8}{3} \times \frac{50}{3}} $

$ M = \sqrt{\frac{400}{9}} $

$ M = \frac{\sqrt{400}}{\sqrt{9}} = \frac{20}{3} $

Thus, the mean proportional is $\frac{20}{3}$.