The average of the first twelve multiples of 11 is:

Correct option is D

Given:
First twelve multiples of 11

Formula Used:

Sum of an Arithmetic Progression (AP) is given by:

​$$S_n = \frac{n}{2} × (2a + (n - 1) d)$$​​

Solution:
First term (a) = 11
Common difference (d) = 22 - 11 = 11
Last term (l) = 132
Find number of terms (n):
l = a + (n - 1) d
132 = 11 + (n - 1) × 11
121 = (n - 1) × 11
n - 1 = 11
n = 12
Now, calculate sum:
S_12 = (12 ÷ 2) × (2 × 11 + (12 - 1) × 11)
= 6 × (22 + 121)
= 6 × 143
= 858
Average = 858 ÷ 12
= 71.5

Alternate Solution:​

Multiples of 11: 11, 22, 33, ..., 132
Sum of first n multiples of a number = number × sum of first n natural numbers
Sum = 11 × (1 + 2 + ... + 12)
= 11 × (12 × 13 ÷ 2)
= 11 × 78
= 858
Average = 858 ÷ 12
= 71.5