Question

Three numbers are in continuous proportion. The middle number is 16 and the sum of other two is 130. Find the smallest among the three numbers

Answer 1

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solution--

means proportion is= 12

sum of = 26

three part of 12  

=12/3= 4

then 26-12= 14

and = 14+4= 18

12-4=8

answer is : 8,12,18

Step-by-step explanation:

Answer 2

Given,

Three numbers are in continuous proportion,

Middle number = 16,

Sum of other 2 numbers = 130.

To find,

Smallest among the 3 numbers.

Solution,

Let the three numbers be a, b, and c. These are said to be in continuous proportion if,

a:b :: b:c, or,

a:b = b:c

This can also be written as,

$\frac{a}{b} =\frac{b}{c}$     ...(1)

Here, it is given that the middle number is 16.

⇒ b = 16

Also, the sum of the other two numbers is 130.

⇒ a + c = 130     ...(2)

From equation (1), we can see that,

ac = b²

Since b = 16,

ac = (16)²

⇒ ac =  256     ...(3)

From equation (2),

c = 130 - a     ...(4)

Substituting this in equation (3), we get,

a(130 - a) = 256

⇒ 130a - a² = 256

⇒ a² - 130a + 256 = 0

Solving the above quadratic equation, we get,

a = 128, and a = 2.

So, from equation (4)

c = 2, and 128

We can choose any one value from the two values of a and c respectively, as both are similar.

So, here, let

a = 128,

c = 2, and

b = 16 (given).

So, the smallest number = c = 2.

Therefore, the smallest number among the three numbers that are in continuous proportion will be 2.