The average of the first nine integral multiples of 3 is
Options:
A . 18
B . 15
C . 12
D . 21
Answer: Option B Answer: (b)Required average = ${3(1+2+3+…+9)}/9$= ${9×10}/{2×3}$ = 15Aliter : Using Rule 9, Here, n = 9, x = 3 Average = x$({1+n}/2)$ = 3$({1+9}/2)$ = 15
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