GRE數(shù)學(xué)考試數(shù)學(xué)概念及解析

第第頁GRE數(shù)學(xué)考試數(shù)學(xué)概念及解析

Averageofnnumbersofarithmeticprogression(AP)istheaverageofthesmallestandthelargestnumberofthem.Theaverageofmnumbercanalsobewrittenas*+d(m-1)/2.

E*ample:

Theaverageofallintegersfrom1to5is(1+5)/2=3

Theaverageofalloddnumbersfrom3to3135is(3+3135)/2=1569

Theaverageofallmultiplesof7from14to126is(14+126)/2=70

remember:

Makesurenonumberismissinginthemiddle.

Withmorenumbers,averageofanascendingAPincreases.

Withmorenumbers,averageofadescendingAPdecreases.

AP:numbersfromsum

giventhesumsofmnumbersofanAPwithconstantincrementd,thenumbersinthesetcanbecalculatedasfollows:

thefirstnumber*=s/m-d(m-1)/2,andthen-thnumberiss/m+d(2n-m-1)/2.

E*ample:

ifthesumof7consecutiveevennumbersis70,thenthefirstnumber*=70/7-2(7-1)/2=10-6=4.thelastnumber(n=m=7)is70/7+2(2*7-7-1)/2=10+6=16.thesetistheevennumbersfrom4to16.

Remember:

giventhefirstnumber*,itiseasytocalculateothernumbersusingtheformulaforn-thnumber:*+(n-1)

AP:numbersfromaverage

allmnumbersofanAPcanbecalculatedfromtheaverage.thefirstnumber*=c-d(m-1)/2,andthen-thnumberisc+d(2n-m-1)/2,wherecistheaverageofmnumbers.

E*ample:

iftheaverageof15consecutiveintegersis20,thenthefirstnumber*=20-1*(15-1)/2=20-7=13andthelastnumber(n=m=15)is20+1*(2*15-15-1)/2=20+7=27.

iftheaverageof33consecutiveoddnumbersis67,thenthefirstnumber*=67-2*(33-1)/2=67-32=35andthelastnumber(n=m=33)is67+2*(2*3