GRE數(shù)學(xué)是怎么樣的呢?又有哪些知識(shí)點(diǎn)呢?下面是小編為大家整理收集的關(guān)于GGRE數(shù)學(xué)重點(diǎn)考察知識(shí)點(diǎn)的相關(guān)解析，希望對(duì)大家有所幫助。

　　Sum of Arithmetic Progression(數(shù)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(guó)數(shù)學(xué)基礎(chǔ)教育的難度增加導(dǎo)致數(shù)學(xué)考試越來(lái)越難，但新gre數(shù)學(xué)復(fù)習(xí)考點(diǎn)都是高中時(shí)候?qū)W到的知識(shí)點(diǎn)，考生不要過(guò)于緊張，把基本概念弄明白，再記住一些新版gre數(shù)學(xué)必備的詞匯，那么相信新版gre數(shù)學(xué)應(yīng)該沒(méi)有問(wèn)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.