Why the Mean Proportional Between 4 and 9 Is 6, Not 6.5

A ratio $a:b$ compares two quantities of the same kind and equals the fraction $\frac{a}{b}$. Four quantities are in proportion, written $a:b=c:d$, when the two ratios are equal. The quick test is cross-multiplication:

$$a:b=c:d⟺ad=bc$$

Here $a$ and $d$ are the extremes, and $b$ and $c$ the means. This part covers special ratios, dividing and changing ratios, proportionals and continued proportion.

Multiply ratios term by term to compound them, square or cube both terms for duplicate or triplicate ratios, and take square or cube roots for sub-duplicate or sub-triplicate ratios.

- Compounded ratio of $a:b$ and $c:d$ is $ac:bd$
- Duplicate ratio of $a:b$ is $a^2:b^2$
- Sub-duplicate ratio is $\sqrt{a}:\sqrt{b}$
- Triplicate ratio is $a^3:b^3$
- Sub-triplicate ratio is $\sqrt[3]{a}:\sqrt[3]{b}$

Worked examples.

$$\text{compounded of }2:3\text{ and }9:16=18:48=3:8$$
$$\text{duplicate of }3:4=9:16\text{sub-duplicate of }49:81=7:9$$
$$\text{triplicate of }2:3=8:27\text{sub-triplicate of }64:125=4:5$$

Combined. Find the compounded ratio of the duplicate of $2:3$, the sub-duplicate of $36:49$ and the triplicate of $7:4$.

$$4:9,6:7,343:64(4\times 6\times 343):(9\times 7\times 64)=8232:4032=49:24$$

An everyday example. **On a map drawn to a scale of $1:1000$**, areas are in the duplicate ratio $1:1000000$.

The substance. Lengths scale by the ratio, areas by its duplicate and volumes by its triplicate.

To divide in a ratio, add the terms to find the total number of parts and share accordingly; when equal amounts are added or removed, write the new ratio as an equation and solve for the unknown.

Worked example 1 — dividing. Share ₹4500 in the ratio $2:3:4$.

$$2+3+4=9\text{ parts}\text{one part}=\frac{4500}{9}=₹500$$

The shares are ₹1000, ₹1500 and ₹2000.

Worked example 2 — adding. What number must be added to each of $7$ and $11$ to make the ratio $3:4$?

$$\frac{7+x}{11+x}=\frac{3}{4}28+4x=33+3xx=5$$

Worked example 3 — subtracting. What number must be taken from each of $23$, $30$, $57$ and $78$ so that they are in proportion?

$$(23-x)(78-x)=(30-x)(57-x)1794-101x=1710-87xx=6$$

Check: $17\times 72=1224=24\times 51$.

An everyday example. **A milk vendor has $40$ L of milk and water in the ratio $5:3$.** That is $25$ L of milk and $15$ L of water, so **$10$ L more water** makes the ratio $1:1$.

The substance. Adding the same amount to both terms changes a ratio, while multiplying both terms by the same number does not.

**The fourth proportional to $a,b,c$ is $\frac{bc}{a}$, the third proportional to $a,b$ is $\frac{b^2}{a}$, and the mean proportional between $a$ and $c$ is $\sqrt{ac}$.

Fourth proportional.** $a:b=c:x$, so $x=\frac{bc}{a}$.

$$\text{fourth proportional to }3,5,12=\frac{5\times 12}{3}=20$$

Third proportional. $a:b=b:x$, so $x=\frac{b^2}{a}$.

$$\text{third proportional to }16,36=\frac{36^2}{16}=\frac{1296}{16}=81$$

Mean proportional. $a:x=x:c$, so $x^2=ac$ and $x=\sqrt{ac}$.

$$\text{mean proportional between }4\text{ and }9=\sqrt{36}=6$$

Algebraic example. Mean proportional between $x-y$ and $x^3-x^{2}y$:

$$\sqrt{(x-y)\cdot x^2(x-y)}=x(x-y)$$

An everyday example. **If $3$ cups of rice serve $5$ people, $12$ cups serve $20$** — the fourth proportional to $3$, $5$ and $12$.

The misconception. The mean proportional is not the average. Between $4$ and $9$ it is $6$, while the average is $6.5$.

**Three quantities $a,b,c$ are in continued proportion when $a:b=b:c$, that is $b^2=ac$; four quantities are in proportion when the product of the extremes equals the product of the means.

Worked example 1.** Are $4,6,9$ in continued proportion? $6^2=36=4\times 9$, so yes.

Worked example 2. Find $x$ if $12,x,27$ are in continued proportion.

$$x^2=12\times 27=324x=18$$

Worked example 3. Test the four quantities.

$$2.5,4,5,8:2.5\times 8=20=4\times 5\text{in proportion}$$
$$3,7,9,20:3\times 20=60\ne 63=7\times 9\text{not in proportion}$$

Worked example 4 — proof. If $a,b,c$ are in continued proportion, show that $(a^2+b^2):(b^2+c^2)=a:c$.

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

An everyday example. The long sides of A3, A4 and A5 paper sheets are in continued proportion, because each sheet is the previous one folded in half with the same shape.

The substance. Order matters: $4,9,6$ are not in continued proportion, even though $4,6,9$ are.