Central Tendency & DispersionPYQ Nov. 20Question 3000 of 411
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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

Options

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

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

All Options:

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

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