IF A.M & G.M of a&b are in ratio m:n Prove
that a:b= m+✓m^2-n^2 : m-✓m²_n²
Answers
Step-by-step explanation:
Given :-
A.M and G.M of a and b are in the ratio is m:n
To find:-
Prove that a:b= m+✓(m^2-n^2) : m-✓(m^2-n^2)
Solution:-
Let a and b be any numbers then
Arithmetic Mean of a and b = (a+b)/2
A.M = (a+b)/2----------(1)
Let a and b be any numbers then
Geometric Mean of a and b = √ab
G.M = √(ab)----------(2)
The ratio of A.M and G.M of a and b =
(a+b)/2 :√ab --------(3)
According to the given problem
The ratio of A.M and G.M of a and b =m:n
=>(a+b)/2 :√ab = m:n
We know tht a:b can be written as a/b
=>[(a+b)/2]/√(ab) = m/n
=>(a+b)/2√ab = m/n
=>a+b = 2√ab ×m/n------(4)
We know that
(a-b)^2 = (a+b)^2-4ab
=>(a-b)^2 = [2√abm/n]^2-4ab
=>(a-b)^2 = 4abm^2/n^2 -4ab
=>(a-b)^2 = (4abm^2-4abn^2)/n^2
=>(a-b)^2 = [4ab(m^2-n^2)]/n^2
=>a-b = √[{4ab(m^2-n^2)}/n^2]
=>a-b = 2√[ab(m^2-n^2)]/n---------(5)
On adding (1)&(2) then
a+b+a-b=[2√ab ×m/n]+2√[ab(m^2-n^2)]/n
=>2a = [2m√ab +2√[ab(m^2-n^2)]]/n
=>2a = 2√ab [m+√(m^2-n^2)/n]
=>a = (√ab/n) [m+(√m^2-n^2)]------(6)
On Substituting the value of a in (4) then
(√ab/n) [m+(√m^2-n^2)] + b = 2m√(ab)/n
=>b= 2m√(ab)/n -[(√ab/n) [m+(√m^2-n^2)]]
=>b =[ 2m√(ab)-[√(ab)[[m+(√m^2-n^2)]]/n
b=[2m√(ab)-√(ab)m -√(ab)(√m^2-n^2)]/n
=>b=[m√(ab)-√(ab)√(m^2-n^2)]/n
=>b = √(ab)/n[m-√(m^2-n^2)]------(7)
Now from (6)&(7)
a:b=>
(√ab/n) [m+(√m^2-n^2)]:(√(ab)/n)[m-√(m^2-n^2)]
On cancelling (√ab/n) then
=>[m+(√m^2-n^2)] : [m-√(m^2-n^2)]
Answer:-
a:b = [m+(√m^2-n^2)] : [m-√(m^2-n^2)]
Used formula:-
- Arithmetic Mean of a and b = (a+b)/2
- Geometric Mean of a and b = √ab
- (a-b)^2 = (a+b)^2-4ab