针对新版gre数学复习,将每天为大家整理一个重要考点,希望大家能够将这些重要考点系统地记住,在考试中取得一个好成绩,突破新gre数学困境。
新版gre数学复习重要考点:Sum of Arithmetic Progression(数列求和)
The sum of n-numbers of an arithmetic progression is given by
S=nxdn(n-1)/2
where x is the first number and d is the constant increment.
example:
sum of first 10 positive odd numbers:101+2109/2=10+90=100
sum of first 10 multiples of 7 starting at 7: 107+7109/2=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(27-7-1)/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(215-15-1)/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(233-33-1)/2=67+32=99.
Remember:
sum of the m numbers is cm,where c is the average.
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