Introduction / Context:
This question checks understanding of averages of terms in an arithmetic progression, specifically numbers divisible by a given integer within a certain range. Instead of listing every number and summing them, we can use properties of arithmetic sequences to work efficiently, which is crucial in competitive exams.
Given Data / Assumptions:
- Range of interest: greater than or equal to 100 and less than or equal to 200.
- We need numbers divisible by 13.
- We seek the average of all such numbers.
- Average of an arithmetic sequence = (first term + last term) / 2.
Concept / Approach:
Numbers divisible by 13 form an arithmetic progression with common difference 13. Within the given range, we find the smallest and largest multiples of 13 that lie between 100 and 200. Once we know the first and last terms of this sequence, the average of all such terms is simply the mean of these two extremes, because the terms are equally spaced.
Step-by-Step Solution:
Find smallest multiple of 13 greater than or equal to 100.
13 * 7 = 91 (too small), 13 * 8 = 104 (acceptable).
So, first term = 104.
Find largest multiple of 13 less than or equal to 200.
13 * 15 = 195 (acceptable), 13 * 16 = 208 (too large).
So, last term = 195.
Average of all terms in this arithmetic sequence = (first term + last term) / 2.
Average = (104 + 195) / 2 = 299 / 2 = 149.5.
Verification / Alternative check:
List the multiples: 104, 117, 130, 143, 156, 169, 182, 195.
There are 8 numbers.
Sum = (104 + 195) * 8 / 2 = 299 * 4 = 1196.
Average = 1196 / 8 = 149.5, which confirms the earlier result.
Why Other Options Are Wrong:
Option A (147.5): Slightly less than the correct value, possibly from arithmetic error.
Option B (145.5): Does not match the mean of the first and last multiples.
Option C (143.5): Too far from the center of the range of multiples.
Option D (149.5): Correct average based on the arithmetic progression formula.
Common Pitfalls:
Students sometimes include 91 or 208 by mistake, which lie outside the given interval.
Another error is averaging 100 and 200 directly instead of focusing on numbers divisible by 13.
Forgetting that the average of equally spaced numbers equals the mean of the first and last terms can also lead to a long and error prone approach.
Final Answer:
The required average of all numbers between 100 and 200 divisible by 13 is 149.5.
Discussion & Comments