If the mean of the first $n$ natural numbers is $\frac{5 n}{9}$, then the value of $n$ is :\

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Professor PC Thomas Classes · Answer

(c) :Mean of first $n$ natural numbers
$ =\frac{1+2+\ldots \ldots+n}{n}=\frac{n(n+1)}{2 n} $ 
$\Rightarrow \quad \frac{n(n+1)}{2 n}=\frac{5 n}{9} \quad\left(ight.$ Given mean $\left.=\frac{5 n}{9}ight)$\ $\Rightarrow 9 n(n+1)=10 n^{2} \Rightarrow 9 n^{2}+9 n=10 n^{2}$\ $\Rightarrow n^{2}-9 n=0 \Rightarrow n(n-9)=0 \Rightarrow n=9$ $$ (\because n 
eq 0) $$