Central Tendency & DispersionMTP Nov 19Question 2883 of 473

The mean of the values of $\displaystyle 1, 2, 3, \dots, n$ with respective frequencies $\displaystyle x, 2x, 3x, \dots, nx$ is.

\displaystyle \frac{n+1}{2}

\displaystyle \frac{n}{2}

\displaystyle \frac{2n+1}{3}

\displaystyle \frac{2n+1}{6}

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Correct Answer

✅ Option c — $\displaystyle \frac{2n+1}{3}$

**Step 1: Set up the weighted mean formula.** Values:

\displaystyle 1, 2, 3, \ldots, n

. Frequencies:

\displaystyle x, 2x, 3x, \ldots, nx

(i.e., frequency of value

\displaystyle r

\displaystyle rx

\bar{x} = \frac{\sum_{r=1}^{n} r \cdot (rx)}{\sum_{r=1}^{n} rx} = \frac{x \sum_{r=1}^{n} r^2}{x \sum_{r=1}^{n} r}

**Step 2: Use standard summation formulas.**

\sum_{r=1}^{n} r = \frac{n(n+1)}{2}, \qquad \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}

**Step 3: Compute the ratio.**

\bar{x} = \frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}} = \frac{n(n+1)(2n+1)}{6} \times \frac{2}{n(n+1)} = \frac{2(2n+1)}{6} = \frac{2n+1}{3}

Hence, **Option C** is the correct answer.

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