Question

A, B, C are three distinct natural numbers less
than 11. The arithmetic mean of A and B is 9 and
the geometric mean of B and C is 672. What are
the values of A, B and C, respectively?​​

Answer 1

Pls check ur question again....

$\sf Arithmetic \: mean \: of \: A \: and \: B \\\\\sf =\frac{A+B} {2}=9 \\\\\sf </p><p>=A+B=18 \\\\\sf </p><p>B=18-A \\\\\sf </p><p>11>A>7</p><p>Geometric \: mean \: of \: B \: and \: C \\\\\sf </p><p>= \sqrt{BxC} =672 \\\\\sf </p><p>=BxC= 672^2 \\\\\sf$

How is it possible.....I think something is wrong in the question bcz if A, B and C are less than 11, then geometric mean of B and C will never be equal to 672.

The maximum value of geometric mean of B and C cannot exceed 11...