Let the two positive numbers be a and b.
It is given that the sum of the two numbers is 6 times their geometric mean.
∴ \[ a + b = 6\sqrt{ab} \]
Squaring both sides, we get \[ (a + b)^2 = 36ab \]
\[ a^2 + b^2 + 2ab = 36ab \]
\[ a^2 + b^2 - 34ab = 0 \quad \text{… (1)} \]
Dividing equation (1) by \( b^2 \), we get \[ \left(\frac{a}{b}\right)^2 - 34\left(\frac{a}{b}\right) + 1 = 0 \]
Let \( \frac{a}{b} = x \). Then, \[ x^2 - 34x + 1 = 0 \]
Solving the quadratic equation: \[ x = \frac{34 \pm \sqrt{34^2 - 4}}{2} \]
\[ x = \frac{34 \pm \sqrt{1156 - 4}}{2} \]
\[ x = \frac{34 \pm \sqrt{1152}}{2} \]
\[ x = \frac{34 \pm 24\sqrt{2}}{2} \]
\[ x = 17 \pm 12\sqrt{2} \]
Therefore, \[ \frac{a}{b} = 17 + 12\sqrt{2} \quad \text{or} \quad 17 - 12\sqrt{2} \]
Dividing numerator and denominator by 1, we write the ratio as \[ a : b = (17 + 12\sqrt{2}) : 1 \]
Taking square roots on both sides of the original ratio form, we get the ratio of the numbers as
\[ (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}) \]
Hence, the two numbers are in the ratio \[ (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}). \]
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab) n .