Re: [討論]問兩題數學
※ 引述《foxiness (北極狐)》之銘言：
: ※ 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
: greatest + 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
: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
不好意思我資質駑鈍..可以解釋或說明一下這個式子怎麼來的嗎?
謝謝!!
= (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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