Question

three numbers are in continued proportion whose mean proportional is 12 and sum of the remaning two numbers is 26 then find these numbers
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Answer 1

Answer:

$assume \: the \: number \: and \: foll$

$let \: the \: three \: numbers \: be \: abc$

as they are in continued proportion

$b {}^{2} = abc$

given that mean proportional is 12

: b=12

so

$12 {}^{2} = ac$

144 = ac

c=144\a

the sum of the two numbers

a+c=26

a+144\a=26

step 2: simplify using factorization

$= a {}^{2} + 144 - 26a = 0 \\ = a { }^{2} - 18a - 8a + 144 = 0 \\ = a(a - 18) - 8(a - 18) = 0 \\ = (a - 18)(a - 8) = 0 \\ = a = 18 \: or \: a = 8 \\ = c = 8 \: or \: c = 18$

a=18 or a=8

c=8 or c=18

hence the numbers are 8,12,18 or 18,12,8

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