Answer :

Let the numbers be x, y, z.

As the numbers are in continued proportion, therefore

y^{2} = 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 – z^{2} = 144

⇒ z^{2} – 26z + 144 = 0

⇒ z^{2} – 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.

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