第2课时等差数列的性质[教材要点]要点一等差数列的性质(1)在等差数列{an}中,p,q,s,t∈N*,且p+q=s+t,则______________.(2)若{an}为公差为d的等差数列,则{can}是公差为cd的等差数列.(3)若{an}为公差为d的等差数列,则{an+an+k}(k∈N*)是公差为2d的等差数列.ap+aq=as+at(4)若{an}是等差数列,公差为d,则{a2n}也是等差数列,公差为________.(5)若{an},{bn}分别是以d1,d2为公差的等差数列,则{pan+pbn}是以________为公差的等差数列.(6)若{an}是等差数列,公差为d,则ak,ak+m,ak+2m,…(k,m∈N*)组成公差为________的等差数列.(7)当d________0时,数列{an}为单调递增数列;当d________0时,数列{an}为单调递减数列;当d=0时,数列{an}为常数列.2dpd1+qd2md><状元随笔若{an}是公差为d的等差数列,则还具有其他性质(1)am+n-an=am+k-ak=md(m,n,k∈N*)(2)下标成等差数列,则数列am,am+k,am+2k,am+3k…成等差数列,公差为kd(m,k∈N*).(3){an}是等差数列,则a1,a3,a5…仍成等差数列(首项不一定选a1).(4)若{bn}为等差数列,则{an±bn},{kan+b}(k,b为非零常数)也为等差数列.(5){an}去掉前几项后余下的项仍组成公差为d的等差数列.(6)奇数项数列{a2n-1}是公差为2d的等差数列;偶数项数列{a2n}是公差为2d的等差数列.(7)若{kn}成等差数列,则{akn}也是等差数列.要点二等差数列的实际应用具有等差特征的实际问题,可构造等差数列模型,用等差数列的知识求解.[基础自测]1.判断正误(正确的画“√”,错误的画“×”)(1)若{an}是等差数列,则{|an|}也是等差数列.()(2)若{|an|}是等差数列,则{an}也是等差数列.()(3)若{an}是等差数列,则对任意n∈N*都有2an-1=an+an+2.()(4)数列{an}的通项公式为an=3n+5,则数列{an}的公差与函数y=3x+5的图象的斜率相等.()××√√2.在等差数列{an}中,a1+a9=10,则a5的值为()A.5B.6C.8D.10解析:由等差数列的性质,得a1+a9=2a5,又 a1+a9=10,即2a5=10,∴a5=5.故选A.答案:A3.在等差数列{an}中,a5=6,a8=15,则a14=()A.21B.32C.33D.42解析:设公差为d,∴a8=a5+(8-5)d,∴d=15-63=3,∴a14=a8+6d=15+6×3=33.故选C.答案:C4.在等差数列{an}中,已知a3+a4+a5+a6+a7=450,则a2+a8=________.解析:因为a3+a4+a5+a6+a7=5a5=450,所以a5=90,a2+a8=2a5=2×90=180.答案:180题型一等差数列的性质应用——师...