新GRE数学重要考点指导

　　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，小编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.