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If AM of numbers $\mathrm{x}_{1}, \mathrm{x}_{2} \ldots . \mathrm{x}_{\mathrm{n}}$ is $\mu$, then what is the $\mathrm{AM}$ of the numbers which are increased by $1,2,3, \ldots \mathrm{n}$ respectively?
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Verified Answer
The correct answer is:
$\mu+\left(\frac{\mathrm{n}+1}{2}\right) \quad$
Since, AM of number $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \ldots \mathrm{x}_{\mathrm{n}}$ is $\mu$
$\mathrm{n} \mu=\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots \mathrm{x}_{\mathrm{n}}$
Sum of new numbers
$=\left(x_{1}+1\right)+\left(x_{2}+2\right)+\left(x_{3}+3\right)+\ldots+\left(x_{n}+n\right)$
$=\left(x_{1}+x_{2}+\ldots+x_{n}\right)+(1+2+3+\ldots+n)$
$=n \mu+\frac{n(n+1)}{2}$
$\mathrm{AM}=\mu+\frac{(\mathrm{n}+1)}{2}$
$\mathrm{n} \mu=\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots \mathrm{x}_{\mathrm{n}}$
Sum of new numbers
$=\left(x_{1}+1\right)+\left(x_{2}+2\right)+\left(x_{3}+3\right)+\ldots+\left(x_{n}+n\right)$
$=\left(x_{1}+x_{2}+\ldots+x_{n}\right)+(1+2+3+\ldots+n)$
$=n \mu+\frac{n(n+1)}{2}$
$\mathrm{AM}=\mu+\frac{(\mathrm{n}+1)}{2}$
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