The average of the first nine integral multiples of 3 is

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The average of the first nine integral multiples of 3 is

1. 21
2. 12
3. 15
4. 18

Correct Option: C

As we know that ,
the first nine integral numbers multiples of 3 are -
3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27
Sum = 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27

Required average =	3(1 + 2 + 3 +...... + 9)
	9

Required average =	9 × 10	= 15
	2 × 3

Second method to find the required average ,
Here, n = 9, y = 3

Average = y		l + n
		2

Average = 3		1 + 9		= 15
		2