Question

Question 19 The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that a:b = [m+(m^2- n^2)^0.5] : [m-(m^2- n^2)^0.5] .

Class X1 - Maths -Sequences and Series Page 200

Answer 1

AM of number a and b = (a + b)/2
GM of number a and b = √ab

Here,
AM : GM = m : n
(a + b)/2 : √ab = m : n
(a + b)/2√ab = m/n
Applying componendo and dividendo rule,
(a + b + 2√ab)/(a + b - 2√ab) = (m + n)/(m - n)
(√a² + √b² + 2√ab)/(√a² + √b² -2√ab) = (m + n)/(m - n)
(√a + √b)²/(√a - √b)² = (m + n)/(m - n)
take square root both sides,
( √a + √b )/(√a - √b) = √(m + n)/√(m - n)
again applying componendo and dividendo,
( √a + √b + √a - √b)/(√a + √b - √a + √b) = {√(m + n) + √(m - n)}/{√(m + n) - √(m - n) }
2√a/2√b = {√(m + n) + √(m - n)}/{√(m + n) - √(m - n) }
√a/√b = {√(m + n) + √(m - n)}/{√(m + n) - √(m - n) }
taking square both sides,
a/b =[{√(m + n) + √(m - n)}/{√(m + n) - √(m - n) }]²
a/b = {m + n + m - n - 2√(m² - n²)}/{m + n + m - n -2√(m² - n²)}
a/b = {2m + 2√(m² - n²)}/{2m - 2√(m² - n²)}
a/b = {m + √(m² - n²)}/{m - √(m² - n²)}

hence, a : b = m + √(m² - n²) : m - √(m² - n²)

Answer 2

the answer is proved

hope it helps you....