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爱尔兰留学 新GRE数学复习重要考点

2014-12-24 10:21:55 来源: 撰稿 :365留学网
 2011新gre考试实施以来,不仅对写作部分茫然不知所措,就连我们所擅长的数学也甚是担忧,相信这其中的原因还是由于考生在新gre数学复习时没有把基础打牢。如果新版gre数学基本考点都没有复习到,如何能拿到分数呢?所以说,想要新版gre数学考好,复习的时候一定要把基本概念和重要考点都弄扎实。

  所以,从今天起,针对新版gre数学复习,每天给考生整理一个重要考点,这些概念在考试中一定会考到的。希望考生能再接再厉,取得一个好成绩,突破新版gre数学难的困境。

  新版gre数学复习重要考点:Sum of Arithmetic Progression

  The sum of n-numbers of an arithmetic progression is given by

  S=nx*dn(n-1)/2

  where x is the first number and d is the constant increment.

  example:

  sum of first 10 positive odd numbers:10*1+2*10*9/2=10+90=100

  sum of first 10 multiples of 7 starting at 7: 10*7+7*10*9/2=70+315=385

  remember:

  For a descending AP the constant difference is negative. 由于美国数学基础教育的难度增加导致数学考试越来越难,但新gre数学复习考点都是高中时候学到的知识点,考生不要过于紧张,把基本概念弄明白,再记住一些新版gre数学必备的词汇,那么相信新版gre数学应该没有问题。

  AP

  Average of n numbers of arithmetic progression (AP) is the average of the smallest and the largest number of them. The average of m number can also be written as x + d(m-1)/2.

  Example:

  The average of all integers from 1 to 5 is (1+5)/2=3

  The average of all odd numbers from 3 to 3135 is (3+3135)/2=1569

  The average of all multiples of 7 from 14 to 126 is (14+126)/2=70

  remember:

  Make sure no number is missing in the middle.

  With more numbers, average of an ascending AP increases.

  With more numbers, average of a descending AP decreases.

  AP:numbers from sum

  given the sum s of m numbers of an AP with constant increment d, the numbers in the set can be calculated as follows:

  the first number x = s/m - d(m-1)/2,and the n-th number is s/m + d(2n-m-1)/2.

  Example:

  if the sum of 7 consecutive even numbers is 70, then the first number x = 70/7 - 2(7-1)/2 = 10 - 6 = 4.

  the last number (n=m=7)is 70/7+2(2*7-7-1)/2=10+6=16.the set is the even numbers from 4 to 16.

  Remember:

  given the first number x, it is easy to calculate other numbers using the formula for n-th number: x+(n-1)

  AP:numbers from average

  all m numbers of an AP can be calculated from the average. the first number x = c-d(m-1)/2, and the n-th number is c+d(2n-m-1)/2, where c is the average of m numbers.

  Example:

  if the average of 15 consecutive integers is 20, then the first number x=20-1*(15-1)/2=20-7=13 and the last number (n=m=15) is 20+1*(2*15-15-1)/2=20+7=27.

  if the average of 33 consecutive odd numbers is 67, then the first number x=67-2*(33-1)/2=67-32=35 and the last number (n=m=33) is 67+2*(2*33-33-1)/2=67+32=99.

  Remember:

  sum of the m numbers is c*m,where c is the average.


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