What is the average of the first 93 natural numbers?

Introduction / Context:
This question tests knowledge of the properties of consecutive natural numbers and the standard formula for their average. The first n natural numbers form an arithmetic progression, so their average is easy to compute using endpoints.

Given Data / Assumptions:
- We consider the first 93 natural numbers: 1, 2, 3, ..., 93. - They form an arithmetic progression with common difference 1. - We must find the average of these 93 numbers.

Concept / Approach:
For any arithmetic progression with first term a and last term l, the average (mean) of all terms is equal to (a + l) / 2. This is because the sequence is symmetric around the middle value. The first n natural numbers fit this pattern with a = 1 and l = n, so the average is (1 + n) / 2.

Step-by-Step Solution:
Step 1: Identify first term a = 1. Step 2: Identify last term l = 93. Step 3: Recognize that the numbers 1 to 93 form an arithmetic progression. Step 4: Use the formula for the average of an arithmetic progression: average = (a + l) / 2. Step 5: Substitute values: average = (1 + 93) / 2. Step 6: 1 + 93 = 94, and 94 / 2 = 47.

Verification / Alternative check:
Note that the middle of 1 and 93 is 47, and the sequence is symmetric: 1 pairs with 93, 2 with 92, and so on, each pair averaging to 47. Since every such pair has average 47, the whole set must also have average 47, confirming the formula based result.

Why Other Options Are Wrong:
- 45, 46, 49 and 50 do not match the midpoint between 1 and 93 and therefore cannot equal the average of this symmetric sequence.

Common Pitfalls:
Some learners try to sum all numbers up to 93 and then divide, which is time consuming and error prone. Others mistakenly think that the average equals the number of terms, 93, or half of 93. Remember the simple midpoint formula (first + last) / 2 for any consecutive sequence.

Final Answer:
The average of the first 93 natural numbers is 47.