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The sum \(\mathrm{S}\) of the arithmetic sequence \(\mathrm{a, a + d, a + 2d, \ldots, a + (n - 1)d}\) is given by \(\mathrm{S_n} = \frac{\mathrm{n}}{2}[2\mathrm{a} + (\mathrm{n}-1)\mathrm{d}]\). What is the sum of the integers from 1 to 100, inclusive, with the even integers between 25 and 63 omitted?
Let's break down what this problem is really asking us to do in plain English.
We want to add up all the whole numbers from 1 to 100. That's straightforward enough - it's like counting 1 + 2 + 3 + 4 + ... all the way up to 100.
But here's the twist: we need to skip certain numbers. Specifically, we need to leave out the even numbers that fall between 25 and 63. Think of it like having a list of numbers 1 through 100, then crossing out some of them before we add up what's left.
The smartest approach is to calculate the full sum first (all numbers 1 to 100), then subtract the numbers we need to remove (the even numbers between 25 and 63).
Process Skill: TRANSLATE - Converting the word problem into a clear mathematical plan
Let's start by finding the sum of all integers from 1 to 100.
This is a classic arithmetic sequence where we start at 1, and each next term increases by 1, until we reach 100. So we have:
- First term (a) = 1
- Common difference (d) = 1
- Number of terms (n) = 100
Using the arithmetic sequence formula given in the problem:
\(\mathrm{S_n = (n/2)[2a + (n-1)d]}\)
Substituting our values:
\(\mathrm{S_{100} = (100/2)[2(1) + (100-1)(1)]}\)
\(\mathrm{S_{100} = 50[2 + 99]}\)
\(\mathrm{S_{100} = 50 \times 101}\)
\(\mathrm{S_{100} = 5,050}\)
So the sum of all integers from 1 to 100 is 5,050.
Now we need to find which even integers fall between 25 and 63, and calculate their sum.
The even integers between 25 and 63 are: 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62
Let's count them: starting from 26 and ending at 62, going up by 2 each time.
This gives us an arithmetic sequence where:
- First term (a) = 26
- Common difference (d) = 2
- Last term = 62
To find the number of terms: \(\mathrm{62 = 26 + (n-1)(2)}\)
\(\mathrm{36 = (n-1)(2)}\)
\(\mathrm{18 = n-1}\)
\(\mathrm{n = 19}\)
So we have 19 even integers to exclude.
Using the arithmetic sequence formula:
\(\mathrm{S_{19} = (19/2)[2(26) + (19-1)(2)]}\)
\(\mathrm{S_{19} = (19/2)[52 + 36]}\)
\(\mathrm{S_{19} = (19/2)(88)}\)
\(\mathrm{S_{19} = 19 \times 44}\)
\(\mathrm{S_{19} = 836}\)
Now we simply subtract the sum of the excluded even integers from our total sum.
Final answer = Sum of 1 to 100 - Sum of excluded even integers
Final answer = 5,050 - 836
Final answer = 4,214
The sum of integers from 1 to 100, with even integers between 25 and 63 omitted, is 4,214.
Looking at our answer choices, this matches option D exactly.
1. Misinterpreting "between 25 and 63"
Students often get confused about whether "between" includes the endpoints. In this problem, "between 25 and 63" means we exclude 25 and 63 themselves, only considering numbers like 26, 28, 30... up to 62. Some students might incorrectly include 24 (the even number before 25) or 64 (the even number after 63).
2. Incorrectly identifying which numbers to exclude
The problem asks to omit "even integers between 25 and 63," but students might misread this as excluding ALL integers between 25 and 63 (both odd and even), or might think they need to exclude even numbers throughout the entire range 1-100. The key is recognizing that we only exclude the even numbers within that specific range.
3. Using the wrong approach entirely
Some students might try to manually list and add all remaining numbers instead of using the more efficient "total sum minus excluded sum" approach. This leads to a much longer calculation with higher chance of arithmetic errors.
1. Arithmetic errors in the sequence formulas
When applying \(\mathrm{S_n = (n/2)[2a + (n-1)d]}\), students commonly make calculation mistakes, especially with the order of operations. For example, when calculating (19/2)(88), students might incorrectly compute this as \(\mathrm{19 \times 44 = 826}\) instead of the correct \(836\).
2. Incorrectly counting the number of terms
When finding how many even numbers exist between 25 and 63, students often miscount. They might forget that to get from 26 to 62 in steps of 2, they need to solve \(\mathrm{62 = 26 + (n-1)(2)}\) correctly, potentially getting n = 18 instead of n = 19.
3. Wrong setup of the excluded sequence
Students might incorrectly identify the first term as 24 or 28 instead of 26, or use the wrong common difference. Since we want even numbers starting after 25, the sequence should begin with 26, not 24 or 28.
1. Calculation verification errors
Students might arrive at the correct approach and setup but make a final arithmetic error when computing 5,050 - 836. They might get 4,224 or 4,204 instead of 4,214, then select the closest answer choice rather than double-checking their calculation.
2. Selecting based on partial calculations
Some students might calculate only the sum 1 to 100 (getting 5,050) and select an answer choice that seems reasonable relative to this number, without properly subtracting the excluded terms. Or they might subtract an incorrect exclusion sum and pick the closest available option.