If the average of m numbers is n^{2} and that | vedclass

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Question

Sum of m numbers $=n^{2} 	imes m$

Sum of n numbers $=m^{2} 	imes n$

Average of $(m+n)=\frac{n^{2} 	imes m+m^{2} 	imes n}{m+n}=\frac{m n(n+m)

Vedclass · Accepted Answer

$m n$

Vedclass · Answer

Sum of m numbers $=n^{2} 	imes m$

Sum of n numbers $=m^{2} 	imes n$

Average of $(m+n)=\frac{n^{2} 	imes m+m^{2} 	imes n}{m+n}=\frac{m n(n+m)}{(n+m)}=m n$

If the average of $m$ numbers is $n^{2}$ and that of $n$ numbers is $m^{2},$ then the average of $( m + n )$ numbers is

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