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The average of thirteen consecutive integers is 36. If two times the smallest of these 13 integers is added to the largest of these 13 integers, what will be the sum obtained?
102
115
121
110
102
For an odd number of consecutive integers, the average is always the middle term. If the average is 36, then 36 is the 7th term. Since there are 13 terms, the smallest is 36тИТ6=30 and the largest is 36+6=42.
There are 13 consecutive integers with an average of 36.
A=2nтАЛ(2a+(nтИТ1)d)
For an odd number of consecutive integers, the average is always the middle term. If the average is 36, then 36 is the 7th term. Since there are 13 terms, the smallest is 36тИТ6=30 and the largest is 36+6=42.
Students often try to solve for n terms using algebra without realizing that for consecutive integers, the average is the median value.
Identify the integers
Since the average of 13 consecutive integers is 36, the middle term (the 7th term) is 36.
Middle=36
Find the smallest and largest integers
There are 6 integers before 36 and 6 integers after 36. Thus, the smallest integer is 36тИТ6=30 and the largest integer is 36+6=42.
Smallest=30,Largest=42
Calculate the final sum
The problem asks for two times the smallest integer added to the largest integer: 2├Ч30+42.
Sum=2(30)+42=60+42=102
A is correct because the smallest integer is 30 and the largest is 42, leading to 2(30)+42=102.
This shortcut applies to any arithmetic sequence with an odd number of terms; the mean is always equal to the median.
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