a5的平方=a10

a5^2=a10.得出(a1*q^4）^2=a1*q^9得出a1=qAn为递增数列,说明q>12[An+A(n+2)]=5A(n+1)A(n+2)=an·q^2;A(n+1)=an·q代入上式得：2A

（x的平方-x+1）^2令x=1a12+a11+a10+a9+a8+a7+a6+a5+a4+a3+a2+a1+a0=1

=a^8(a²+a-1)+a^5(a²+a-1)+a^2(a²+a-1)=(a^8+a^5+a^2)(a²+a-1)=0

a3*a9=a4*a8=4a6*a10=a8²a3*a5=a4²所以a4*a8=4a8²+a4²=41a4+a8>0(a4+a8)²=a4²

答：设a1=x,等差为y.则a1+a2+...+a5=30为5x+10y=30;a6+a7+...+a10=80为5x+35y=80;解得x=2y=2;a11+a12+...a15则为5x+60y=1

在等差数列{an}中,a1＋a10＝a2＋a9＝a3＋a8＝a4＋a7＝a5＋a6＝2∴a1－a2＋a3－a4＋a5＋a6－a7＋a8－a9＋a10＝(a1＋a10)＋(a3＋a8)＋(a5＋a6)－

设a1=m,a2=km,a3=k^2m,a4=k^3m…an=k^(n-1)ma3*a4*a5=k^2*k^3*k^4m^3=k^9m^3=3a6*a7*a8=k^5*k^6*k^7m^3=k^18m

=SUM(INDIRECT("a"&ROW(a1)*3-2):INDIRECT("a"&ROW(a1)*3))为了达到效果是：b2=sum(a1:a3),b3=sum(a4:a6),b4=sum(a7

设公比为q(a9+a10+a11+a12)/(a5+a6+a7+a8)=16/1(a5q⁴+a6q⁴+a7q⁴+a8q⁴)/(a5+a6+a7+a8)=

a2a10=a5a7=6又a2+a10=5故a2a10是x²-5x+6=0的解(x-2)(x-3)=0解得x=2或3若a2=2,a10=3则a18/a10=a10/a2=3/2若a2=3,a

公比为qa5.a9.a13.a17=81a5.a5q^4.a5q^8.a5q^12=a5^4q^24=81a11=a5q^6=3a4.a7=a1q^3.a10/q^3=a1.a10

a2=a1+(2-1)d=a1+d=5(1)a5=a1+(5-1)d=a1+4d=14(2)(2)-(1)得到3d=9d=3,a1=2所以a10=a1+(10-1)d=2+9*3=29

4因为a3=a1+2d（d为等差）a4=a1+3da5=a1+4da7=a1+6da10=a1+9da13=a1+12d全部换过来的话3(a3+a5)+2(a7+a10+a13)=24=》a1+6d=

在等差数列{an}中若m+n=k+l则am+an=ak+al因为2（a3+a5）+2a10=4所以由等差数列上述性质得：a4+a10=a1+a13=2．所以S13=13×(a1+a13)2＝13．故选

=SUM(A2:A10)这是从A2单元格到A10单元格求和；SUM(A2:A5,A6:A10)这是从A2单元格到A5单元格求和,然后再从A6到A10单元格求和；从这里看,结果是一样的；不同的是,在EX

因为数列为等差数列由a4+a5+a6+a7++a14=77得11a9=77∴a9=7它又由a4+a7+a10=18得3a7=6②由①②可得首项a1=3,d=1/2由等差数列通项公式an=a1+(n-1

2a2=2(1+d)a10=1+9d5a5=5(1+4d)它们成等比,所以(1+9d)^2=5(1+4d)×2(1+d)解得d=82/81或-2/9(舍去)所以an=a1+(n-1)×d=1-82/8

｛an｝各项为整数,所以d为整数,且d≠0a5=6→a2=6-3da10=6+5da2*a10=(6-3d)(6+5d)>10→36+2d-15d^2>10→15d^2-2d-260d=-1时,a2=

a2=5a5=11,设公差为da5-a2=3d=6所以d=2a10=a5+5d=21