Question

Three numbers are in continued proportion, whose mean proportional is 12 and the

sum of the remaining 2 numbers is 26, then find these numbers​

Answer 1

Solution :

Let the numbers in continued proportion be a and b .

Their mean proportional is 12 .

Hence , the proportion is -

> a :: 12 :: 12 :: b

Now , the sum of these numbers is 26 .

Hence , a + b = 26

Also , the product of them , ab = 12 × 12 = 144

ab = 144

> b = 144/a

Placing this in the first equation

> a + 144/a = 26

> a² + 144 = 26a

> a² - 26a + 144 = 0

> a² - 18a - 8 a + 144 = 0

> a( a - 18) - 8( a - 18) = 0

> ( a - 8)( a - 18) = 0

Hence , there are two values of a that is 8 and 18

If a is 8, b is 144/a = 18 and the opposite for a = 18 .

Answer - The two numbers are tuples of ( 8, 18)

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Answer 2

Question:-

• Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining 2 numbers is 26, then find these numbers.

To Find:-

• Find the numbers.

Solution:-

Given ,

• Three numbers are in continued proportion.

• Their mean is 12.

Then ,

• $\sf \: \dfrac { a } { 12 } = \dfrac { 12 } { b }$

• $\sf \: a + b = 26$ . . . . ( 1 )

• $\sf \: ab = 144$ . . . . ( 2 )

• $\sf \: b = \dfrac { 144 } { a }$

Now ,

Substitute b in equation ( 1 ):-

$\sf\longrightarrow \: a + \dfrac { 144 } { a } = 26$

$\sf\longrightarrow \: \dfrac{ { a }^{ 2 } + 144 } { a } = 26$

$\sf\longrightarrow \: { a }^{ 2 } - 26a + 144$

$\sf\longrightarrow \: { a }^{ 2 } - 18a - 8a + 144 = 0$

$\sf\longrightarrow \: a( a - 18 ) - 8( a - 18 ) = 0$

$\sf\longrightarrow \: ( a - 18 ) ( a - 8 ) = 0$

$\sf\longrightarrow \: a = 18 , 8$

Now ,

We have to find the value of ‘ b ’ of a = 18:-

$\sf\longrightarrow \: a + b = 26$

$\sf\longrightarrow \: 18 + b = 26$

$\sf\longrightarrow \: b = 26 - 18$

$\sf\longrightarrow \: b = 8$

Hence ,

• Numbers are 18 , 8