If the A.M and G.M of two numbers are 13 and 12 respectively then the two numbers are 

Correct option is (C) 8, 18

Let both numbers are a and b.

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

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

\(\therefore\) \(\frac{a+b}2=13\)

And \((ab)^\frac12=12\)              (Given)

\(\Rightarrow a+b=26\)          _______________(1)

& \(ab=12^2=144\)    _______________(2)

\(\therefore(a-b)^2=(a+b)^2-4ab\)

\(=(28)^2-4\times144\)

\(=676-576\)

\(=100=10^2\)

\(\Rightarrow a-b=10\)       _______________(3)

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

\((a+b)+(a-b)=26+10\)

\(\Rightarrow2a=36\)

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

\(\therefore b=26-a\)

\(=26-18=8\)           (From (1))

Hence, required numbers are 8 and 18.

Correct option is C) 8, 18