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Q.

Find the average of cubes of first 24 natural numbers

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a

60,000

b

80,000

c

70,000

d

90,000

answer is B.

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Detailed Solution

Average of first ‘n’ perfect cubes

$= {[\frac{n (n + 1)}{2}]}^{2}$

$\Rightarrow {(\frac{24 (24 + 1)}{2})}^{2}$

$\Rightarrow {(\frac{24 \times 25}{2})}^{2}$

$= 300^{2} = 90, 000$

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