The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio .
Let the two numbers be a and b.
G.M. = √ab
According to the given condition,
a + b = 6√ab
squaring on both sides we get,
⇒ (a + b)2 = 36(ab) —1
Here,
(a—b)2 = (a + b)2 – 4ab = 36ab – 4ab = 32ab
⇒ a—b = √32√ab = 4√2√ab —2
Adding eq(1) and eq(2), we get
2a = (6 + 4√2)√ab
⇒ a = (3 + 2√2)√ab
Substituting the value of a in eq(1), we obtain
b = 6√ab – (3 + 2√2)√ab
⇒ b = (3 — 2√2)√ab
Now,
∴ The required ratio is