# Three numbers are

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.

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