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Zorluk: ZorConsecutive Integers and Number Sets

Set SS consists of nn consecutive integers, where n>1n > 1. The sum of all the elements in Set SS except the greatest element is 360360, and the sum of all the elements in Set SS except the least element is 440440. What is the median of the elements in Set SS?

  1. A
    00
  2. B
    44
  3. 55Cevap
  4. D
    88
  5. E
    1010

Cevap

The median of the elements in Set SS is 55.
Subtracting the given partial sums gives the difference between the largest and smallest elements: (T−a1)−(T−an)=440−360=80(T - a_1) - (T - a_n) = 440 - 360 = 80. For a set of nn consecutive integers, an−a1=n−1a_n - a_1 = n - 1, so n=81n = 81. In any set of consecutive integers, the arithmetic mean equals the median, mm. Therefore, the total sum of all 8181 elements is 81m81m. The middle term is the 41st element (mm), which means the 81st element is m+40m + 40. Substituting these into T−an=360T - a_n = 360 yields 81m−(m+40)=36081m - (m + 40) = 360, simplifying to 80m=40080m = 400, which gives m=5m = 5.

Adım Adım Çözüm

1
Set up equations for the total sum TT of Set SS.
Let a1a_1 be the least element and ana_n be the greatest element. T−an=360T - a_n = 360 and T−a1=440T - a_1 = 440.
Subtracting the greatest element leaves 360360, and subtracting the least element leaves 440440.
2
Find the difference between the greatest and least elements an−a1a_n - a_1.
(T−a1)−(T−an)=440−360  ⟹  an−a1=80(T - a_1) - (T - a_n) = 440 - 360 \implies a_n - a_1 = 80.
Subtracting the two sum equations eliminates the total sum TT.
3
Determine the number of elements nn in Set SS.
For consecutive integers, an−a1=n−1a_n - a_1 = n - 1. Thus, n−1=80  ⟹  n=81n - 1 = 80 \implies n = 81.
The difference between the nn-th and 1st term of consecutive integers is n−1n - 1.
4
Relate the total sum TT and the greatest element ana_n to the median mm.
Since n=81n = 81 is odd, the mean equals the median mm. Total sum T=81mT = 81m. The greatest element is a81=m+40a_{81} = m + 40.
In an evenly spaced set, total sum is n×mn \times m, and the last term is m+n−12m + \frac{n-1}{2}.
5
Solve for the median mm.
T−an=360  ⟹  81m−(m+40)=360  ⟹  80m−40=360  ⟹  80m=400  ⟹  m=5T - a_n = 360 \implies 81m - (m + 40) = 360 \implies 80m - 40 = 360 \implies 80m = 400 \implies m = 5.
Substituting T=81mT = 81m and an=m+40a_n = m + 40 into the first equation allows solving for mm directly.

Anahtar Kavram

Mean-Median Equivalence and Counting Terms in Consecutive Integer Sets
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Konular

Consecutive Integers and Number Sets

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