The mean of n observations is \(\bar{x}\). If three observations \(\mathrm{n}+1, \mathrm{n}-1,2 \mathrm{n}-1\) are added such that mean remains same, then value of n is

Let the n observations be \(x_1, x_2, \ldots, x_{\mathrm{n}}\)
\(\therefore \quad\) According to have given condition; we get
\(\begin{aligned}
& \frac{x_1+x_2+\ldots+x_n}{n} \\
& =\frac{x_1+x_2+\ldots+x_n+\mathrm{n}+1+\mathrm{n}-1+2 \mathrm{n}-1}{\mathrm{n}+3} \\
\therefore \quad & \bar{x}=\frac{x_1+x_2+\ldots+x_n+4 \mathrm{n}-1}{\mathrm{n}+3} \\
\therefore \quad & \bar{x}=\frac{\left(\frac{x_1+x_2+\ldots+x_n}{n}\right)+4 \mathrm{n}-1}{\mathrm{n}+3}
\end{aligned}\)
\(\begin{array}{ll}\therefore & \bar{x}=\frac{\mathrm{n} \bar{x}+4 \mathrm{n}-1}{\mathrm{n}+3} \\ \therefore & \quad \mathrm{n} \bar{x}+3 \bar{x}=\mathrm{n} \bar{x}+4 \mathrm{n}-1 \\ \therefore & 3 \bar{x}=4 \mathrm{n}-1 \\ \therefore & \mathrm{n}=\frac{3 \bar{x}+1}{4}\end{array}\)