The average of a non-decreasing sequence of N numbers a₁,a₂,…,aₙ is 300. If a₁ is replaced by 6a₁, the new average becomes 400. Then, the number of possible values of a₁ is
Question:
The average of a non-decreasing sequence of N numbers a₁,a₂,…,aₙ is 300. If a₁ is replaced by 6a₁, the new average becomes 400. Then, the number of possible values of a₁ is
The average of a non-decreasing sequence of N numbers a₁,a₂,…,aₙ is 300. If a₁ is replaced by 6a₁, the new average becomes 400. Then, the number of possible values of a₁ is
Options
Answer: (2) 14
Explanation:
Let the sequence length be N (a positive integer) and the numbers be a1 ≤ a2 ≤ ... ≤ aN. Initial average = 300, so total sum S = 300N. After replacing a1 by 6a1, new total = S - a1 + 6a1 = S + 5a1. New average = (S + 5a1)/N = 400. Using S = 300N: (300N + 5a1)/N = 400 ⇒ 300 + 5a1/N = 400 ⇒ 5a1/N = 100 ⇒ a1 = 20N. Feasibility condition (since sequence is non-decreasing): each of a2,...,aN ≥ a1, so the minimum possible sum of the last N−1 terms is (N−1)a1. Thus S = 300N ≥ a1 + (N−1)a1 = N·a1. Substitute a1 = 20N: 300N ≥ N·(20N) ⇒ 300 ≥ 20N ⇒ N ≤ 15. Also N must be at least 2 (N = 1 would force a1 = 20 but the initial average is 300, contradiction). So valid integer N are 2,3,...,15. Each such N gives a distinct a1 = 20N. Therefore the number of possible values of a1 is 14. Answer: 14
Explanation:
Let the sequence length be N (a positive integer) and the numbers be a1 ≤ a2 ≤ ... ≤ aN. Initial average = 300, so total sum S = 300N. After replacing a1 by 6a1, new total = S - a1 + 6a1 = S + 5a1. New average = (S + 5a1)/N = 400. Using S = 300N: (300N + 5a1)/N = 400 ⇒ 300 + 5a1/N = 400 ⇒ 5a1/N = 100 ⇒ a1 = 20N. Feasibility condition (since sequence is non-decreasing): each of a2,...,aN ≥ a1, so the minimum possible sum of the last N−1 terms is (N−1)a1. Thus S = 300N ≥ a1 + (N−1)a1 = N·a1. Substitute a1 = 20N: 300N ≥ N·(20N) ⇒ 300 ≥ 20N ⇒ N ≤ 15. Also N must be at least 2 (N = 1 would force a1 = 20 but the initial average is 300, contradiction). So valid integer N are 2,3,...,15. Each such N gives a distinct a1 = 20N. Therefore the number of possible values of a1 is 14. Answer: 14
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