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.
