If the arithmetic mean of two numbers is 10 and geometric mean is 6, then the numbers are 

Correct option is (B) 2, 18

Let both required numbers are a and b.

\(\therefore\) Their arithmetic mean \(=\frac{a+b}2\)

and their geometric mean \(=(ab)^\frac12\)

\(\therefore\frac{a+b}2=10\;\&\;(ab)^\frac12=6\)          (Given)

\(\Rightarrow a+b=20\)      _______________(1)

& \(ab=6^2=36\)   _______________(2)

Now, \((a-b)^2=(a+b)^2-4ab\)

\(=20^2-4\times36\)

\(=400-144\)

\(=256=16^2\)

\(\therefore a-b=16\)     _______________(3)

By adding (1) & (3), we get

\((a+b)+(a-b)=20+16\)

\(\Rightarrow2a=36\)

\(\Rightarrow a=\frac{36}2=18\)

\(\therefore b=20-a\)

\(=20-18=2\)    (From (1))

Hence, both numbers are 2 and 18.

Correct option is B) 2, 18