Given the weights for the numbers 1, 2, 3, dots, n are respectively 1^2, 2^2, 3^2, dots, n^2 then weighted HM is

Option c: 2n+1/3. Solution: Step 1: Recall the formula for Weighted Harmonic Mean. text{Weighted HM} = sum wi/sum frac{wi{xi}} where xi = i and wi = i^2. Step 2: Compute numerator. sum{i=1}^{n} wi = sum{i=1}^{n} i^2 = n(n+1)(2n+1)/6 Step 3: Compute denominator. sum{i=1}^{n} wi/xi = sum{i=1}^{n} i^2/i = sum{i=1}^{n} i = n(n+1)/2 Step 4: Calculate Weighted HM. text{WHM} = dfrac{n(n+1)(2n+1)/6}{dfrac{n(n+1)}{2}} = n(n+1)(2n+1)/6 x 2/n(n+1) = 2n+1/3 Hence, Option C is the correct answer.

Central Tendency & DispersionPYQ Nov. 20Question 3000 of 473

All Questions

Given the weights for the numbers $\displaystyle 1, 2, 3, \dots, n$ are respectively $\displaystyle 1^2, 2^2, 3^2, \dots, n^2$ then weighted HM is

Options

\displaystyle \frac{2n+1}{4}

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

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

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

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Correct Answer

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

All Options:

A $\displaystyle \frac{2n+1}{4}$
B $\displaystyle \frac{2n+1}{6}$
C $\displaystyle \frac{2n+1}{3}$
D $\displaystyle \frac{2n+1}{2}$

Detailed Solution & Explanation

**Step 1: Recall the formula for Weighted Harmonic Mean.**

\text{Weighted HM} = \frac{\sum w_i}{\sum \frac{w_i}{x_i}}

where

\displaystyle x_i = i

and

\displaystyle w_i = i^2

. **Step 2: Compute numerator.**

\sum_{i=1}^{n} w_i = \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}

**Step 3: Compute denominator.**

\sum_{i=1}^{n} \frac{w_i}{x_i} = \sum_{i=1}^{n} \frac{i^2}{i} = \sum_{i=1}^{n} i = \frac{n(n+1)}{2}

**Step 4: Calculate Weighted HM.**

\text{WHM} = \frac{\dfrac{n(n+1)(2n+1)}{6}}{\dfrac{n(n+1)}{2}} = \frac{n(n+1)(2n+1)}{6} \times \frac{2}{n(n+1)} = \frac{2n+1}{3}

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

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Given the weights for the numbers 1,2,3,…,n\displaystyle 1, 2, 3, \dots, n1,2,3,…,n are respectively 12,22,32,…,n2\displaystyle 1^2, 2^2, 3^2, \dots, n^212,22,32,…,n2 then weighted HM is