Central Tendency & DispersionMTP Dec 2023 Series IIQuestion 2925 of 473
All Questions

The weighted mean of first n\displaystyle n natural numbers, if their weights are proportional to their corresponding numbers is

Options

A2n+13\displaystyle \frac{2n+1}{3}
Bn−12\displaystyle \frac{n-1}{2}
Cn(n+1)(2n−1)6\displaystyle \frac{n(n+1)(2n-1)}{6}
D3n(n+1)2\displaystyle \frac{3n(n+1)}{2}
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Correct Answer

✅ Option a — 2n+13\displaystyle \frac{2n+1}{3}

All Options:

  • A2n+13\displaystyle \frac{2n+1}{3}
  • Bn−12\displaystyle \frac{n-1}{2}
  • Cn(n+1)(2n−1)6\displaystyle \frac{n(n+1)(2n-1)}{6}
  • D3n(n+1)2\displaystyle \frac{3n(n+1)}{2}

Detailed Solution & Explanation

**Step 1: Set up the weighted mean.** Values: 1,2,3,…,n\displaystyle 1, 2, 3, \ldots, n. Weights proportional to the numbers themselves: wi=i\displaystyle w_i = i. xˉw=∑i=1ni⋅wi∑i=1nwi=∑i=1ni⋅i∑i=1ni=∑i=1ni2∑i=1ni\bar{x}_w = \frac{\sum_{i=1}^{n} i \cdot w_i}{\sum_{i=1}^{n} w_i} = \frac{\sum_{i=1}^{n} i \cdot i}{\sum_{i=1}^{n} i} = \frac{\sum_{i=1}^{n} i^2}{\sum_{i=1}^{n} i} **Step 2: Use standard formulas.** ∑i=1ni=n(n+1)2,∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i = \frac{n(n+1)}{2}, \quad \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} xˉw=n(n+1)(2n+1)6n(n+1)2=n(n+1)(2n+1)6×2n(n+1)=2n+13\bar{x}_w = \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{2n+1}{3} **Note:** The correct answer is 2n+13\displaystyle \frac{2n+1}{3} (Option A), not Option C. The `correct_option` given as C is n(n+1)(2n−1)6\displaystyle \frac{n(n+1)(2n-1)}{6} which is not the weighted mean formula result. Hence, **Option A** is the correct answer.

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