设等差数列an的公差为d,若a1a3=4-2d2,则4a2 a5 a7的最小值为
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![设等差数列an的公差为d,若a1a3=4-2d2,则4a2 a5 a7的最小值为](/uploads/image/f/7262009-17-9.jpg?t=%E8%AE%BE%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E7%9A%84%E5%85%AC%E5%B7%AE%E4%B8%BAd%2C%E8%8B%A5a1a3%3D4-2d2%2C%E5%88%994a2+a5+a7%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA)
a100=-1+(100-1)×(-3)=-298a1=-1a2=-1-3=-4a3=-4-3=-7a4=-7-3=-10a5=-10-3=-13
由题意得:a11=a1+(11-1)d=a1+10d=0所以a1=-10dS14=(a1+a14)14/2=7(a1+a14)=98所以a1+a14=14又因为a14=a1+13d=3d所以-10d+
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
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
设等差数列的公差为d,则a3=a5-2d=6-2d,an1=a5+(n1-5)d=6+(n1-5)d.∵a3,a5,an1成等比数列,∴a52=a3an1化简即(6n1-42)d-2(n1-5)d2=
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
设An=A1+(n-1)dBm=B1*q^(m-1)(此处楼上打错)因为A1=B1;A3=B3;A7=B5则可得A1(1-q^2)=2dA1(1-q^4)=6d比得q^4-3q^2+2=0(q^2-1
设a2=k,a1=k-a,a3=k+ab1*b3=b2*b2解出来k=-a/3b1=-a/3b2=2a/3b3=-4a/3公比q=-2再问:a是什么?
等差数列所以a1a2a3a4a5a6a7平均数是a4则方差=[(a1-a4)^2+……+(a7-a4)^2]/7=1(a1-a4)^2=(-3d)^2=9d^2(a2-a4)^2=(-2d)^2=4d
S2-S1=(an+1-a1)+(an+2-a2)+...+(a2n-an)=nd*n=d*n^2S3-S2=(a2n+1-a1)+(a2n+2-a2)+...+(a3n-a2n)=nd*n=d*n^
题目都打错了,这是江苏今年的高考题.
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:
n=9再问:为什么再答:等一下,我写一下过程再答:再答:再答:不懂追问再问:那你上面为什么说n=9?再答:说错了再答:n=9和10时结果一样再答:n=10时an=0再问:哦哦,我知道了
∵数列{an}是等差数列∴a1+a2+a3+a4+a5+a6+a7=7a4;则a1,a2,a3,a4,a5,a6,a7的平均数为a4,∴a1,a2,a3,a4,a5,a6,a7的方差为17[(a1-a
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
再问:太给力了,你的回答完美解决了我的问题!
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.