Math, asked by prafulldwivedi2003, 1 month ago

If the arithmetic mean of two numbers is
3 times their geometric means. Then
find the ratio of those numbers.​

Answers

Answered by jainnikansh106
0

Answer:

Let the numbers be a and b.

Given

2

a+b

=3

ab

b

a

+

a

b

=6

⇒t+

t

1

=6 (where t=

b

a

)

⇒t

2

−6t+1=0

⇒t=

2

6

2

−4

⇒t=3±2

2

b

a

=17±12

2

⇒(i)

b

a

=17+12

2

∴ option C is correct.

or (ii)

b

a

=17−12

2

a

b

=

17−12

2

1

=

(3−2

2

)

2

1

∴ option

Answered by amitnrw
0

Given  Arithmetic mean of two numbers is 3 times their geometric means.  

To find : the ratio of those numbers.​

Solution:

Let say Two numbers are a and ka

Hence ratio is 1 : k

arithmetic mean =  (a + ka)/2  = a(k + 1)/2

Geometric mean = √a.ak = a√k

arithmetic mean of two numbers is 3 times their geometric means.

=> a(k + 1)/2 = 3a√k

=> k + 1 = 6√k

Squaring both sides

=> k² + 2k + 1 = 36k

=> k² -34k + 1 =  0

=> k = (34 ± √(-34)² - 4.1.1)/2

=> k = 17 ± √17² - 1

=> k = 17 ± √288

=> k = 17 ± 12√2

Ratio of numbers  is

1 : (17 +  12√2)   or  1  : (17  - 12√2)

1 : (3 + 2√2)²     or   1  : (3 - 2√2)²

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