List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers

题目

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?

选项

解析

对于任何等距集合，**中位数 = 平均数 = 首项与末项的平均值**。集合\(S\)的平均数是首项与末项的平均值：设首项为\(x\)，则末项为\(x + 9×2\)，因此平均数 = \(\frac{x + (x + 9×2)}{2} = x + 9\)；集合\(T\)的平均数就是其中位数（即第3项）：设\(T\)的首项为\(x - 7\)，则第3项为\((x - 7) + 2×2 = x - 3\)，因此平均数 = \(x - 3\)；两者的差值为：\((x + 9) - (x - 3) = 12\) 答案：D

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