The average of the first nine integral multiples of 3 is:
Answer & explanation
Correct answer: option 3
3 6 9 -----------------------------27
Average = \(\frac{1st\;+\;Last}{2}\) = \(\frac{3\;+\;27}{2}\) ⇒ 15
The average of the first nine integral multiples of 3 is:
Correct answer: option 3
3 6 9 -----------------------------27
Average = \(\frac{1st\;+\;Last}{2}\) = \(\frac{3\;+\;27}{2}\) ⇒ 15