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