Question

If three quantities are in continued proportion, show that the ratio of the first to the third is the duplicate ratio of the first to the second.

(That is, if the three quantities are x, y, and z such that x : y = y : z, prove that \[x : z = x^2 : y^2\].)

Answer 1

Given: Three quantities x, y, and z are in continued proportion. x : y = y : z \[\Rightarrow \frac{x}{y} = \frac{y}{z}\]

To Prove: The ratio of the first to the third is the duplicate ratio of the first to the second, i.e., \[x : z = x^2 : y^2\]

Proof: Since x, y, and z are in continued proportion: \[\frac{x}{y} = \frac{y}{z}\] \[\Rightarrow y^2 = xz\]

By definition, the duplicate ratio of the first quantity to the second quantity is: \[\text{Duplicate ratio of } x : y = x^2 : y^2\] \[= \frac{x^2}{y^2}\]

Substituting \[y^2 = xz\] into the denominator: \[= \frac{x^2}{xz}\] \[= \frac{x}{z}\] \[= x : z\]

\[\therefore x : z = x^2 : y^2\]

Hence, the ratio of the first quantity to the third quantity is equal to the duplicate ratio of the first to the second. Hence Proved.

Alternative Method (k-method):

Let: \[\frac{x}{y} = \frac{y}{z} = k\]

\[\Rightarrow y = zk\]

\[\Rightarrow x = yk = (zk)k = zk^2\]

Ratio of the first to the third: \[\text{L.H.S.} = \frac{x}{z}\]

\[= \frac{zk^2}{z}\] 

\[= k^2\]

Duplicate ratio of the first to the second: \[\text{R.H.S.} = \frac{x^2}{y^2}\]

\[= \frac{(zk^2)^2}{(zk)^2}\]

\[= \frac{z^2 k^4}{z^2 k^2}\]

\[= k^2\]

Since \[\text{L.H.S.} = \text{R.H.S.}\]: \[x : z = x^2 : y^2\]

Hence Proved.