Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.

Let the numbers be x, y, z.

As the numbers are in continued proportion, therefore

y² = xz ……………… (1)

Also, the mean proportion = 12

∴ y = √xz = 12

⇒ xz = 144 …………… (2)

It is given that the sum of remaining two numbers = 26

∴ x + z = 26

⇒ x = 26 – z

Put the value of x in equation (2):

(26 – z)z = 144

⇒ 26z – z² = 144

⇒ z² – 26z + 144 = 0

⇒ z² – 8z – 18z + 144 = 0

⇒ z(z - 8) – 18(z – 8) = 0

⇒ (z - 8)(z - 18) = 0

⇒ z = 8 or z = 18

∴ x = 26 – 8 or x = 26 – 18

⇒ x = 18 or x = 8

y = 12

∴ The numbers in proportion be 8, 12, 18 or 18, 12, 8.