Question

3. Three numbers are in continued proportion whose
mean proportional is 12 and the sum of the remain-
ing two numbers is 26. Find the numbers.​

Answer 1

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$\bold {Solution:}$

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

Then

$b = 12 \: and \: a + c = 26 \\ {b}^{2} = ac \: \: \: \: \: \: \: {∴12}^{2} = ac \\ ∴ac = 144 \: \: \: \: \: \: \: \: \: \: \: \: ∴a = \frac{144}{c}$

Substituting a = $\frac{144}{c}$ from (2) in a + c = 26 in (1),

$\frac{144}{c}$ + c = 26

∴144 + $\sqrt {c}^{2}$ = 26c

....(multiplying both sides by c )

∴ $\sqrt {c}{2}$ - 26 + 144 = 0

∴$\sqrt {c}{2}$ - 18c - 8c + 144 = 0

∴ c(c-18)-8(c-18)=0

∴ (c-18) (c-8) = 0

∴c - 18 = 0 or c - 8 = 0

∴ c = 18 or c = 8

Now, a + c = 26 ∴ a + 18 = 26

∴ a = 26 - 18 ∴ a = 8

OR a + c = 26 ∴ a + 8 = 26

∴ a = 26 - 8 ∴ a = 18

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