If a and b are two real numbers such that 0 < a < b and the arithmetic

1.	If a and b are two real numbers such that 0 < a < b and the arithmetic mean between a and b is \(\frac{4}{3}\) times the harmonic mean between them, then \(\frac{b}{a}\) is equal to(a) \(\frac{2}{\sqrt3}\)(b) \(\frac{3}{2}\)(c) 3 (d) \(\frac{8}{3}\)
Answer» (c) 3 Arithmetic mean between a and b is A = \(\frac{a+b}{2}\) Harmonic mean between a and b is H = \(\frac{2ab}{a+b}\) Given, A = \(\frac{4}{3}\)H ⇒ \(\frac{a+b}{2}\) = \(\frac{4}{3}\)\(\big(\frac{2ab}{a+b}\big)\) ⇒ 3(a + b)² = 16ab ⇒ 3b² – 10ab + 3a²= 0 ⇒ 3\(\big(\frac{b}{a}\big)^2\) - 10\(\big(\frac{b}{a}\big)\) + 3 = 0 (Dividing throughout by a²) ⇒ \(\big(\frac{3b}{a}-1\big)\)\(\big(\frac{b}{a}-3\big)\) = 0 ⇒ \(\frac{b}{a}\) = \(\frac{1}{3}\) or \(\frac{b}{a}\) = 3 \(\bigg[\because\,0<a<b\implies\frac{b}{a}\neq\frac{1}{3}\bigg]\) ⇒ \(\frac{b}{a}\) = 3.

If a and b are two real numbers such that 0 < a < b and the arithmetic mean between a and b is \(\frac{4}{3}\) times the harmonic mean between them, then \(\frac{b}{a}\) is equal to(a) \(\frac{2}{\sqrt3}\)(b) \(\frac{3}{2}\)(c) 3 (d) \(\frac{8}{3}\)

Answer»

(c) 3

Arithmetic mean between a and b is A = \(\frac{a+b}{2}\)

Harmonic mean between a and b is H = \(\frac{2ab}{a+b}\)

Given, A = \(\frac{4}{3}\)H ⇒ \(\frac{a+b}{2}\) = \(\frac{4}{3}\)\(\big(\frac{2ab}{a+b}\big)\)

⇒ 3(a + b)² = 16ab ⇒ 3b² – 10ab + 3a²= 0

⇒ 3\(\big(\frac{b}{a}\big)^2\) - 10\(\big(\frac{b}{a}\big)\) + 3 = 0 (Dividing throughout by a²)

⇒ \(\big(\frac{3b}{a}-1\big)\)\(\big(\frac{b}{a}-3\big)\) = 0

⇒ \(\frac{b}{a}\) = \(\frac{1}{3}\) or \(\frac{b}{a}\) = 3 \(\bigg[\because\,0<a<b\implies\frac{b}{a}\neq\frac{1}{3}\bigg]\)

⇒ \(\frac{b}{a}\) = 3.