Central Tendency & DispersionMTP Dec 2023 Series IIQuestion 2925 of 454
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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}
Bn12\displaystyle \frac{n-1}{2}
Cn(n+1)(2n1)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 cn(n+1)(2n1)6\displaystyle \frac{n(n+1)(2n-1)}{6}

All Options:

  • A2n+13\displaystyle \frac{2n+1}{3}
  • Bn12\displaystyle \frac{n-1}{2}
  • Cn(n+1)(2n1)6\displaystyle \frac{n(n+1)(2n-1)}{6}
  • D3n(n+1)2\displaystyle \frac{3n(n+1)}{2}

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