GRE數(shù)學(xué)考試數(shù)學(xué)概念及解析

GRE數(shù)學(xué)考試數(shù)學(xué)概念及解析，很多同學(xué)對(duì)于這個(gè)問題有疑問和不解，那么下面就跟著中國教育在線的小編詳細(xì)了解一下吧。

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.

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