If the average of three consecutive numbers is 42, find the second smallest of these numbers.

Correct option is D

Given:

Average of three consecutive numbers = 42

Concept Used:

Average = $$\frac{\text{Sum of numbers}}{\text{Number of numbers}}$$​​

Consecutive numbers differ by 1.

Solution:

Let the three consecutive numbers be n, n+1, and n+2.

Average = $$\frac{n + (n+1) + (n+2)}{3}$$=42

3n + 3 = 42 $$\times$$​ 3

3n + 3 = 126

3n = 123

n = 41

The numbers are 41, 42, and 43.

Therefore, the second smallest number is 42.

Option (d) is right.