Question

Three numbers are in continued proportion. Their sum is 38 and the product of first and third is 144 find the numbers.​

Answer 1

Let the number,

a , b , c

So,

$\frac{b}{a} = \frac{c}{b}$

$= > {b}^{2} = ac$

According to the QUESTION

$a + b + c = 38$

and

$a \times c = 144$

So,

${b}^{2} = 144$

$= > b = 12$

So,

$a + c + 12 = 38$

$= > a + c = 38 - 12$

$= > a + c = 26$

Now,

$a + c = 26$

$ac = 144$

Then,

$a - c = \sqrt{ {a + b}^{2} - 4ac }$

$= > \sqrt{ {a + c}^{2} - 4ac }$

$= > \sqrt{ {26}^{2} - 4 \times 144 }$

$= > \sqrt{676 - 576}$

$= > \sqrt{100}$

$= > 10$

Now,

$a + c = 26 \\ \\ a - c = 10 \\ \\ + \: \: \: \: \: + \\ \\ = > 2a = 36$

$= > a = 18$

$= > c = 8$

So,

$a = 18$

$b = 12$

$c = 8$

Answer 2

Step-by-step explanation:

1. a=18
2. b=12
3. c=8

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