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Zorluk: OrtaConsecutive Integers and Number Sets

A sequence consists of kk consecutive positive odd integers. The arithmetic mean of the 33 largest integers in the sequence is 2929. If the sum of all kk integers in the sequence is 207207, what is the value of kk?

  1. A
    7
  2. 9Cevap
  3. C
    11
  4. D
    15
  5. E
    23

Cevap

The total number of terms in the sequence, kk, is 9.
The arithmetic mean of 3 consecutive odd integers is their middle term, so the three largest terms are 27, 29, and 31. The largest term is 31. Using the sum formula for an arithmetic progression, S=ka+312=207S = k \cdot \frac{a + 31}{2} = 207. Expressing the first term as a=312(k1)=332ka = 31 - 2(k - 1) = 33 - 2k yields k(32k)=207k(32 - k) = 207, which simplifies to k232k+207=0k^2 - 32k + 207 = 0. The roots are k=9k = 9 and k=23k = 23. Because all integers in the sequence must be positive, a=332k>0a = 33 - 2k > 0, requiring k16k \le 16. Therefore, k=9k = 9.

Adım Adım Çözüm

1
Determine the largest integer in the sequence.
The largest integer is 3131.
For any 3 consecutive odd integers, the arithmetic mean is equal to the middle integer. Since the mean is 2929, the three largest integers are 27,29,3127, 29, 31, so the maximum term is 3131.
2
Express the smallest term aa in terms of kk.
a=332ka = 33 - 2k.
The kk-th term of a consecutive odd integer sequence starting at aa is given by 31=a+2(k1)    a=332k31 = a + 2(k - 1) \implies a = 33 - 2k.
3
Apply the positivity constraint.
k16k \le 16.
Since all terms are positive integers, the smallest term must satisfy a1    332k1    k16a \ge 1 \implies 33 - 2k \ge 1 \implies k \le 16.
4
Set up and solve the sum equation for kk.
k=9k = 9.
The sum of an arithmetic sequence is S=k×a+L2S = k \times \frac{a + L}{2}. Substituting S=207S = 207, L=31L = 31, and a=332ka = 33 - 2k gives 207=k×(332k)+312=k(32k)    k232k+207=0207 = k \times \frac{(33 - 2k) + 31}{2} = k(32 - k) \implies k^2 - 32k + 207 = 0. Factoring yields (k9)(k23)=0(k - 9)(k - 23) = 0. Since k16k \le 16, k=9k = 9.

Anahtar Kavram

The sum of a sequence of consecutive evenly-spaced numbers equals the number of terms multiplied by the average of the first and last terms.
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