NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.Misc.Q.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 + √ (m2 – n2)]: [m + √ (m2 – n2)]
Let the two numbers be a and b.
A.M = (a + b) ÷ 2 and GM = √ab
According to the given condition,
(a + b) ÷ 2√ab = m ÷ n
(a + b)2 ÷ 4ab = m2 ÷ n2
(a + b)2 = 4abm2 ÷ n2
(a + b) = 2m√(ab) ÷ n ……………. (1)
Using this in the identity (a – b)2 = (a + b)2 – 4ab, we obtain
(a – b)2 = (4abm2 ÷ n2) – 4ab
= 4ab (m2 - n2) ÷ n2
a – b = 2√(ab) √ (m2 - n2) ÷ n ………… (2)
Adding equation (1) and (2), we obtain
2a = [2√(ab) {m + √ (m2 - n2)}] ÷ n
a = [√(ab) {m + √ (m2 - n2)}] ÷ n
Substitute the value of a in equation (1),
we get
b = (2m√(ab) ÷ n) – [[√(ab) {m + √ (m2 - n2)}] ÷ n]
b = (m√(ab) ÷ n) – [√(ab) × √ (m2 - n2)}] ÷ n
b = {√(ab) ÷ n} [m - √ (m2 - n2)]
Now, a ÷ b = [[√(ab) {m + √ (m2 - n2)}] ÷ n] ÷ [{√(ab) ÷ n} [m - √ (m2 - n2)]]
a ÷ b = [m + √ (m2 - n2)] ÷ [m - √ (m2 - n2)]
a: b = [m + √ (m2 - n2)]: [m - √ (m2 - n2)]