设an是一个公差为d 的等差数列若1 a1a2=1 a2a3 1 a3a4

a2=a1+d a4=a1+3d（a2）2=a1×a4即（a1+d）2=a1（a1+3d）整理得a1d=d2∵d≠0∴a1=dS10=10a1+12×10×9×d=10a1+45d=55a1

由AK是A1与A2k的等比中项,得得(AK)^2=A1*A2K因为A1=9d所以AK=8+KdA2K=8+2Kd所以（8+Kd)^2=9d*(8+2Kd)(K-4)*(k+2)=0因为K>0所以K=4

首先算这几个数的平均数,易得平均数就是a4.然后跟据方差的计算公式s=1/7[(a1-a4)^2+(a2-a4)^2+(a3-a4)^2+(a4-a4)^2+(a5-a4)^2+(a6-a4)^2+(

a1^2=a11^2,∴a1=-a11a1=-(a1+10d)2a1=-10da1=-5dan=a1+(n-1)d=-5d+(n-1)d=(n-6)d∵d0,a6=0,a7

a1=9dak=a1+(k-1)*d=9d+(k-1)*da2k=a1+(2k-1)*d=9d+(2k-1)*dak^2=a1*a2k化简后可求出k=4

a2+a4=2*a3=8a3=4,a4=3因此a1=6,d=-1通项为an=6-(n-1)=7-n

（1）由等差数列的通项公式及求和公式可得a1+2d+a1+4d=220a1+20×19d2=150∴d=1，a1=-2（2）∵bn=2an-2an+1=21-n=(12)n-1∴bnbn-1=12∴数

设{an}是一个公差为d(d≠0)的等差数列,它的前10项s10=110且a1,a2,a4成等比数列.a1*a4=a2^2a1*(a1+3d)=(a1+d)^2a1=d或d=0(舍去）an=d*nsn

a1=9d则ak=9d+(k-1)d,a2k=9d+(2k-1)d因为ak为a1和ak的等比中项则有ak的平方等于a1乘以a2k即｛9d+(k-1)d}^2=9d{9d+(2k-1)d}化简消去d得：

设a1为a,公差为d,根据等比数列性质,a1*a4=a2^2a(a+3d)=(a+d)^2,得a=d又S10=10a+10*(10-1)d/2=110,a=d=2所以等差数列an=2n只会第一问,第2

A2=A1+dA4=A1+3d(A2)^2=A1×A4(A1+d)^2=A1(A1+3d)(A1)^2+2A1d+d^2=(A1)^2+3A1dA1d=d^2d≠0A1=dS10=10A1+(1/2)

证：a1,a2,a4成等比数列,则a2²=a1a4(a1+d)²=a1(a1+3d)整理,得d²-a1d=0d(d-a1)=0d≠0,要等式成立,只有d-a1=0a1=d

设{an}是一个公差为d(d≠0)的等差数列,它的前10项s10=110且a1,a2,a4成等比数列.a1*a4=a2^2a1*(a1+3d)=(a1+d)^2a1=d或d=0(舍去）an=d*nsn

由等差数列有：S10=10a1+45d=165由等比数列有：(a1+d)(a1+d)=a1(a1+3d)即是a1^2+2d*a1+d^2=a1^2+a1*3dd^2=a1*d所以a1=d；代入第一式:

A2*A2=A1*A4A2=A1+dA4=A1+d得A1=dA10=10dS10=10(A1+A10)/2=110A1=d=2An=2n

（I）由a1，a2，a4成等比数列可得：(a1+2)2＝a1(6+a1)∴4=2a1即a1=2∴an=2+2（n-1）=2n（II）∵bn=n•2an，=n•22n=n•4n∴Sn＝1•4+2•42+

ak=a1+（k-1）d=9d+（k-1）d=（k+8）da2k=a1+（2k-1）d=9d+（2k-1）d=（2k+8）d又a1a2k=ak^2,即9d（8+2k）d=[（8+k）d]^2k=4

设该等差数列是首项为a1,公差为dS3=3a1+3(3-1)*d/2=3a1+3dS2=2a1+2(2-1)*d/2=2a1+dS4=4a1+4(4-1)*d/2=4a1+6d又:S3²=9

a2^2=a1a4(a1+d)^2=a1(a1+3d)a1^2+2a1d+d^2=a1^2+3a1da1d=d^2a1=da1=da2=d+d=2da3=d+2d=3d.an=a1+(n-1)d=d+

因为ak是a1与a2k的等比中项，则ak2=a1a2k，[9d+（k-1）d]2=9d•[9d+（2k-1）d]，又d≠0，则k2-2k-8=0，k=4或k=-2（舍去）．故选B．