Re: [討論]問兩題數學
※ 引述《CSL (原諒我的無能為力)》之銘言：
: 1.If y is the average(arithmetic mean) of
: n consecutive positive integers, n>1
: what is the sum of the greatest and least of
: these integers?
: 答案是2y
n consecutive positive integers:
a, a+1, a+2, a+3,....,a+k (n = k+1)
y = ( a + a+1 + a+2 + a+3 + ... + a+k ) / (k+1)
= ( a(k+1) + k(1+k)/2) / (k+1)
= a + k/2
the sum of the greatest and least = a + a + k = 2a + k = 2(a+k/2) = 2y
: 2. S is a set of n consecutive integers
: The mean of S The median of S
: 答案是C一樣大
: 謝謝!!
n consecutive positive integers:
a, a+1, a+2, a+3,....,a+k (n = k+1)
(1) Assume k = odd n = k+1 = even
The mean of S = ( a + a+1 + a+2 + a+3 + ... + a+k ) / (k+1)
= a + k/2
The median of S = ((a+(k-1)/2) + (a+(k+1)/2)) /2
= (2a + k)/2
= a + k/2
(median:偶數個數就是兩個中間數的平均數)
(2) Asuume k = even n = k+1 = odd
The mean of S = ( a + a+1 + a+2 + a+3 + ... + a+k ) / (k+1)
= a + k/2
The median of S = a + k/2
(median:奇數個數就是中間那個數)
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◆ From: 125.225.96.2
※ 編輯: foxiness 來自: 125.225.97.58 (10/27 16:32)
※ 編輯: foxiness 來自: 125.225.97.58 (10/27 16:33)
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