Concept of Continued Proportion

Three numbers are in continued proportion if:

First:Second = Second:Third

a:b = b:c, then a, b, and c are in continued proportion.

• b is the mean proportional between a and c.

• c is the third proportional to a and b.

a : b = b : c

\[\frac{a}{b}\] = \[\frac{b}{c}\]            

⇒ b × b = a × c

⇒ `(b^2)` = ac and

b =   \[\sqrt{ac}\]

Find the mean proportional between

\[\frac{1}{18}\] and \[\frac{1}{8}\]

Solution: Let the required mean proportional be x. 
\[\frac{1}{18}\], x, and  \[\frac{1}{8}\]  are in continued proportion.

⇒  \[\frac{1}{18}\]: x : \[\frac{1}{18}\]

⇒ x × x =  \[\frac{1}{18}\] × \[\frac{1}{8}\] i.e., x² = \[\frac{1}{144}\]

⇒ x = \[\sqrt{\frac{1}{144}}\] = \[\frac{1}{12}\]

∴ Required mean proportional = \[\frac{1}{12}\]

2.8 and 0.7

Solution:
Let the required mean proportional be x.
∴ 2.8, x and 0.7 are in continued proportion

⇒2.8: x = x:0.7

⇒ x × x = 2.8 × 0.7, i.e., x² = 1.96

⇒ x =  \[\sqrt{1.96}\] = 1.4

∴ Required third proportional = 1.4

₹15 and ₹45

Solution:
Let the required third proportional be ₹x.
∴ ₹15, ₹45 and ₹x are in continued proportion

That is, 15:45 = 45:x

15x = 45 × 45, i.e., x = 135

∴ Required third proportional = ₹135 

16 and 36. 

Solution:
Let the required third proportional be x.
∴ 16, 36 and x are in continued proportion

⇒ 16 : 36 = 36 : x

i.e., 16x = 36 × 36

⇒ x = `"36 × 36" / 16` = 81

∴ Required third proportional = 81

• Continued proportion compares three numbers. a:b = b:c

• Mean proportional (middle): b =  \[\sqrt{ac}\]

• Third proportional (last): c =   `(b^2)/a`

• Used for patterns, sharing, and scaling in real life and science.