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