The first 10 odd numbers starting from 1 are 1, 3, 5, ..., 19. By what percentage is their average less than the last term 19?

Introduction / Context:
This question deals with averages of arithmetic sequences and percentage comparison. We are given the first ten odd numbers starting from 1 and asked to compare their arithmetic mean with the last term of the sequence, which is 19. The phrase by what percentage is the average less than the last term means we must compute the difference between the last term and the average, then express that difference as a percentage of the last term itself.

Given Data / Assumptions:
- The first 10 odd numbers starting from 1 form an arithmetic progression: 1, 3, 5, ..., 19.
- The first term is 1, the common difference is 2, and the 10th term is 19.
- The arithmetic mean of a sequence is (sum of terms) / (number of terms). For an arithmetic progression, the mean equals the average of the first and last terms.
- We interpret the required percentage as (difference / last term) × 100.

Concept / Approach:
We first find the average of the 10 odd numbers using the simple property of arithmetic progressions: mean = (first term + last term) / 2. Then we compute how much smaller this average is than the last term by taking last term minus average. Finally, to express this as a percentage, we divide that difference by the last term and multiply by 100. The result will be a fraction that we can match against the given answer forms.

Step-by-Step Solution:
Step 1: List the first 10 odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. Step 2: Note that this is an arithmetic progression with first term a = 1 and last term l = 19. Step 3: The average (mean) of an arithmetic progression is (a + l) / 2. Thus mean = (1 + 19) / 2 = 20 / 2 = 10. Step 4: The last term is 19, so the average is less than the last term by 19 - 10 = 9. Step 5: Required percentage = (difference / last term) × 100 = (9 / 19) × 100. Step 6: Write this as a single fraction: 9 × 100 / 19 = 900 / 19 percent.

Verification / Alternative check:
We can verify by computing the sum of the sequence directly: sum S = n/2 × (first + last) = 10/2 × (1 + 19) = 5 × 20 = 100. The mean is S / n = 100 / 10 = 10, which agrees with the earlier calculation. The last term is still 19, and the difference is 9. Dividing 9 by 19 and multiplying by 100 again yields 900 / 19 percent. This consistency confirms the calculation is correct.

Why Other Options Are Wrong:
Options 991/16 %, 874/13 %, 719/17 %: None of these fractions simplifies to 9 × 100 / 19. They correspond to very different decimal values and are not obtained by the required difference over the last term.
Option None of these: Incorrect, because 900/19 % is exactly one of the options and matches the correct calculation.

Common Pitfalls:
A common mistake is to compute the percentage with respect to the average instead of the last term, leading to (9 / 10) × 100 = 90 percent. Another error is to misinterpret the question as asking by what percentage the average is greater instead of less, which reverses the sign conceptually. Carefully reading that we want the percentage by which the average is less than the last term and using difference / last term × 100 avoids such confusion.

Final Answer:
The average is less than the last term by 900/19 percent.