Average of first 'N' multiples of 3 is 66. What is the value of 'N'?

Correct option is D

Given:
Average of first 'N' multiples of 3 is 66.
Formula Used:
Sum of first N integers = $$\frac{N(N + 1)}{2}$$​​
Average = $$\frac{Sum}{N}$$​​
Solution:
The first N multiples of 3 are 3, 6, 9, ..., 3N.
Their sum can be written by factoring out 3.
Sum = 3 × (1 + 2 + 3 + ... + N)
Sum = $$\frac{3N(N + 1)}{2}$$​​
To find the average, divide the sum by N.
Average = $$\frac{3(N + 1)}{2}$$​​
We are given that the average is 66.
$$\frac{3(N + 1)}{2}$$​ = 66
3(N + 1) = 132
N + 1 = 44
N = 43
Final Answer
So the correct answer is (d)