新GRE數(shù)學(xué)重要考點(diǎn)解析之平均數(shù)-智課教育旗下智課教育_第1頁
新GRE數(shù)學(xué)重要考點(diǎn)解析之平均數(shù)-智課教育旗下智課教育_第2頁
新GRE數(shù)學(xué)重要考點(diǎn)解析之平均數(shù)-智課教育旗下智課教育_第3頁
新GRE數(shù)學(xué)重要考點(diǎn)解析之平均數(shù)-智課教育旗下智課教育_第4頁
全文預(yù)覽已結(jié)束

下載本文檔

版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請(qǐng)進(jìn)行舉報(bào)或認(rèn)領(lǐng)

文檔簡(jiǎn)介

1、 智 課 網(wǎng) G R E 備 考 資 料 新GRE數(shù)學(xué)重要考點(diǎn)解析之平均數(shù)-智課教育旗下智課教 育新GRE數(shù)學(xué)重要考點(diǎn)解析,本文就平均數(shù)有關(guān)的新GRE數(shù)學(xué)重要考 點(diǎn)做解析,希望可以供考生參考,更好地準(zhǔn)備GRE考試,并最終取得理 想的考生成績(jī)。本文就平均數(shù)有關(guān)的GRE數(shù)學(xué)考試重點(diǎn)做解析,希望可以幫助大家 鞏固基礎(chǔ),更好地備考GRE數(shù)學(xué)。Average of n numbers of arithmetic progression (AP is the average of the smallest and the largest number of them. The average of m n

2、umber can also be written as x + d(m-1/2.Example:The average of all integers from 1 to 5 is (1+5/2=3The average of all odd numbers from 3 to 3135 is(3+3135/2=1569The average of all multiples of 7 from 14 to 126 is(14+126/2=70remember:Make sure no number is missing in the middle.With more numbers, av

3、erage of an ascending AP increases. With more numbers, average of a descending AP decreases. AP:numbers from sumgiven 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

4、-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=7is 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

5、 n-th number: x+(n-1AP:numbers from averageall 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 isc+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 t

溫馨提示

  • 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請(qǐng)下載最新的WinRAR軟件解壓。
  • 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請(qǐng)聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
  • 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁內(nèi)容里面會(huì)有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
  • 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
  • 5. 人人文庫網(wǎng)僅提供信息存儲(chǔ)空間,僅對(duì)用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對(duì)用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對(duì)任何下載內(nèi)容負(fù)責(zé)。
  • 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請(qǐng)與我們聯(lián)系,我們立即糾正。
  • 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對(duì)自己和他人造成任何形式的傷害或損失。

最新文檔

評(píng)論

0/150

提交評(píng)論