Find the weighted arithmetic mean of the first 'n' natural numbers, the weights being the corresponding numbers.

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SSC CGL 2022 Tier-I Official Paper (Held On : 02 Dec 2022 Shift 2)
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  1. \(\rm \frac{{\{ n(n + 1)(2n + 1)\} }}{6}\)
  2. \(\rm \frac{{\{ n(n + 1)\} }}{2}\)
  3. \(\rm \frac{{(2n + 1)}}{3}\)
  4. n

Answer (Detailed Solution Below)

Option 3 : \(\rm \frac{{(2n + 1)}}{3}\)
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Detailed Solution

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Concept used:

1. Sum of the squares of first n natural numbers = \(\frac {n(n +1)(2n +1)}{6}\)

2. Sum of n natural numbers = \(\frac {n(n +1)}{2}\)

Calculation:

First n natural numbers = 1, 2, 3, ...., n

Their corresponding weights respectively = 1, 2, 3, ...., n

Now, the weighted arithmetic mean of the first 'n' natural numbers

⇒ \(\frac {1 \times 1 + 2 \times 2 +3 \times 3 + ..... + n \times n}{1 + 2 + 3 + .... + n}\)

⇒ \(\frac {1^2 + 2^2 + 3^2 + .... + n^2}{1 + 2 + 3 + .... + n}\)

⇒ \(\frac {\frac {n(n +1)(2n +1)}{6}} {\frac {n(n +1)}{2}}\)

⇒ \(\rm \frac{{(2n + 1)}}{3}\)

∴ The weighted arithmetic mean of the first 'n' natural numbers is \(\rm \frac{{(2n + 1)}}{3}\).

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