Difficulty: Easy
Correct Answer: 28
Explanation:
Introduction / Context:
This question asks about the average of the first several multiples of a given number, in this case 7. The multiples of any fixed number form an arithmetic progression. Knowing this allows you to use a simple formula or property to find the average without adding all the terms one by one.
Given Data / Assumptions:
Concept / Approach:
The multiples of 7 form an arithmetic progression (AP) with first term a = 7 and common difference d = 7. In any AP, the average of all its terms is equal to the average of the first and last terms, that is, (first term + last term) / 2. This shortcut saves time compared with adding all the terms. Alternatively, we could add the terms and divide by 7 to double check.
Step-by-Step Solution:
Step 1: Write the first seven multiples of 7 explicitly. 7, 14, 21, 28, 35, 42, 49. Step 2: Use the AP average property. First term = 7, last term = 49. Average = (first + last) / 2 = (7 + 49) / 2 = 56 / 2 = 28. So the required average is 28.
Verification / Alternative Check:
We can verify by direct summation: 7 + 14 + 21 + 28 + 35 + 42 + 49. Pairing terms symmetrically, (7 + 49) + (14 + 42) + (21 + 35) + 28 = 56 + 56 + 56 + 28 = 196. Dividing this total by 7 terms gives 196 / 7 = 28, which agrees with our shortcut method.
Why Other Options Are Wrong:
7 and 14 are just single multiples; they are not central enough in the list of seven multiples to represent the overall average.
21 is one of the interior terms but is below the true central tendency of the full seven-term sequence.
Common Pitfalls:
Some learners mistakenly think the average of the first n multiples of 7 is always 7n / 2, which is not correct without checking the sequence. Others add the numbers but make arithmetic mistakes. Remember the simple rule: for any arithmetic progression, the average equals (first term + last term) / 2, which here simplifies the work greatly.
Final Answer:
The average of the first seven multiples of 7 is 28.
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