GRE數學是怎么樣的呢?又有哪些知識點呢?下面是小編為大家整理收集的關于GGRE數學重點考察知識點的相關解析，希望對大家有所幫助。
Sum of Arithmetic Progression(數列求和)
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.
由于美國數學基礎教育的難度增加導致數學考試越來越難，但新gre數學復習考點都是高中時候學到的知識點，考生不要過于緊張，把基本概念弄明白，再記住一些新版gre數學必備的詞匯，那么相信新版gre數學應該沒有問題。
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.