Central Tendency & DispersionPYQ Nov. 20Question 3000 of 454
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Given the weights for the numbers 1,2,3,,n\displaystyle 1, 2, 3, \dots, n are respectively 12,22,32,,n2\displaystyle 1^2, 2^2, 3^2, \dots, n^2 then weighted HM is

Options

A2n+14\displaystyle \frac{2n+1}{4}
B2n+16\displaystyle \frac{2n+1}{6}
C2n+13\displaystyle \frac{2n+1}{3}
D2n+12\displaystyle \frac{2n+1}{2}
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Correct Answer

Option c2n+13\displaystyle \frac{2n+1}{3}

All Options:

  • A2n+14\displaystyle \frac{2n+1}{4}
  • B2n+16\displaystyle \frac{2n+1}{6}
  • C2n+13\displaystyle \frac{2n+1}{3}
  • D2n+12\displaystyle \frac{2n+1}{2}

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