The sum of the A.M. and G.M. of two positive numbers is equal to the difference between the numbers. The numbers are in the ratio

Question

The sum of the A.M. and G.M. of two positive numbers is equal to the difference between the numbers. The numbers are in the ratio (A) 1: 3 (B) 1: 6 (C)  9: 1 (D) 1: 12

Tardigrade Admin · Accepted Answer

(a+b/2) +√ab = a-b ⇒ 2√ab = a-3b ⇒ 4ab = (a-3b)2 ⇒ a2 -10ab +9b2 = 0 ⇒ (a-9b)(a-b) = 0 ⇒ a= 9b [∵ a≠ b] ⇒ (a/b) = (9/1)

Q. The sum of the $A . M$ . and $G . M$ . of two positive numbers is equal to the difference between the numbers. The numbers are in the ratio

$1 : 3$

$1 : 6$

$9 : 1$

$1 : 12$

$\frac{a + b}{2} + ab = a - b$
$\Rightarrow 2 ab = a - 3 b$
$\Rightarrow 4 ab = (a - 3 b)^{2}$
$\Rightarrow a^{2} - 10 ab + 9 b^{2} = 0$
$\Rightarrow (a - 9 b) (a - b) = 0$
$\Rightarrow a = 9 b [∵ a \neq = b]$
$\Rightarrow \frac{a}{b} = \frac{9}{1}$

Q. The sum of the A.M. and G.M. of two positive numbers is equal to the difference between the numbers. The numbers are in the ratio

1:3

1:6

9:1

1:12