Question

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

Answer 1

x:12::12:y is proportion

x/12=12/y

xy=144

x+y=26

so by

(x-y)^2=(x+y)^2-4xy,

x-y=10

x=18, y=8

Answer 2

Answer:

The required numbers are 8, 12 and 18 or 18, 12 and 8

Step-by-step explanation:

Let a, b and c be the three numbers in continued proportion.

Then b = 12 and a + c = 26 ...(1)

b² = ac

=> 12² = ac

=> ac = 144

=> a = 144/c (2)

Substituting a = 144/c from (2) in a + c in (1)

=> 144/c + c = 26

=> 144 + c² = 26c

=> c² - 18c + 8c + 144 = 0

=> c(c - 18) + 8(c - 18) = 0

=> (c - 18) (c + 8) = 0

=> c - 18 = 0

=> c = 18

OR

=> c - 8 = 0

=> c = 8

Now, a + c = 26

=> a = 26 - 18

=> a = 26 - 8

OR, a + c = 26

=> a = 26 - 8

=> a = 18

The required number 8, 12 and 18 or 18, 12 and 8.

Hope it helps :)