新GRE数学考点：AP.

2017-08-09 作者： 311阅读

　　今天针对新gre数学复习，小编给大家整理的是关于AP的相关内容，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gre数学难的困境。

　　获得更多留学知识请咨询澳际免费在线顾问

　　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.

　　俗话说：“工欲善其事必先利其器”。有一部分考生认为新GRE数学具有一定的难度，这其中的原因是由于考生在复习时没有把新gre数学的基础打牢。想要新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。