设an是公差不为零的等差数列,满足a4² a5²
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![设an是公差不为零的等差数列,满足a4² a5²](/uploads/image/f/7246172-20-2.jpg?t=%E8%AE%BEan%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%8D%E4%B8%BA%E9%9B%B6%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%BB%A1%E8%B6%B3a4%C2%B2+a5%C2%B2)
由于为等比数列,只要连续3项就可确定数列的首项和公比!故只需要讨论4项删去某一项后剩3项即可!故只要讨论a1,a2,a3,a4即可!(1)删掉首项:a2,a3,a4a3^2=(a3-d)(a3+d)d
(1)根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项:an=n(2)根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0
a9=a5+4da15=a5+10d(a5+4d)²=a5(a5+10d)8da5+16d²=10da516d²-2da5=02d(8d-a5)=0d=a5/8所以a9=
1.若n=4时,则原数列为a1,a2,a3,a4.⑴若删去a1,则a3∧2=a2×a4,→d=0,矛盾⑵若删去a2,→a5=0矛盾⑶若删去a3→a1=d→a1/d=1⑷若删去a4→d=0矛盾综上所述,
(1)∵数列{an}是公差不为零的等差数列,a1=2,且a2,a4,a8成等比数列,∴(2+3d)2=(2+d)(2+7d),解得d=2,∴an=2n.(2)∵an=2n,∴3an=32n=9n,此数
用求和公式,求解二元二次方程组.
(1)a3=a1+2d、a6=a1+5d.(a1+2d)^2=a1(a1+5d)a1^2+4a1d+4d^2=a1^2+5a1d4a1d+4d^2=5a1d因为d0,所以4a1+4d=5a1a1=4d
(1)由题意可得(a1+d)2+(a1+2d) 2=(a1+3d)2+(a1+4d)27a1+21d=7联立可得a1=-5,d=2∴an=-5+(n-1)×2=2n-7,sn=−5n+n(n
由题意,显然该等比数列的公比不会是负数,也不会是小于一的数.前者不会满足等差数列要求,后者末项趋于零,不合理.故公比大于一,故等差数列是递增的即公差大于0.又a5*a5=a3*an1即36=a3*an
首项为a1,公差为dS10=10a1+45d=110.(1)a1,a2,a4成等比数列.(a2)^2=a1*a4(a1+d)^2=a1(a1+3d).(2)通过(1)(2)得a1=d=2an=a1+(
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
把首项和公差设出来解个二元一次方程组就行了设首项为a1公差为d则(1)[a1+(a1+d)+(a1+2d)]^2=9[a1+(a1+d)](2)a1+(a1+d)+(a1+2d)+(a1+3d)=4[
在等差数列中,公差d不为0,a11+40d=a51,即a11=a51-40d因为|a11|=|a51|,即a11=-a51,或者a11=a51(不符,舍去)所以a11+a51=2*a31=0,即a31
(1)因为a4,a5,a8成等比数列,所以a52=a4a8.设数列{an}的公差为d,则(3+3d)2=(3+2d)(3+6d)化简整理得d2+2d=0.∵d≠0,∴d=-2.于是an=a2+(n-2
解a1=1a2=1+da5=1+4da1a2a5成等比所以(1+d)^2=1*(1+4d)d^2-2d=0d=2d=0(舍)所以an=a1+(n-1)d=1+(n-1)*2=2n-1
a1a2a3成等比数列a2^2=a1a3=a3(a1+d)^2=a1+2da1^2+2a1d+d^2=a1+2d1+2d+d^2=1+2dd^2=0d=0公差不为零的等差数列错题
数列{an}是公差不为0的等差数列,设公差为d,S1,S2,S4成等比数列,则S22=S1•S4,∴( 2a1+d)2=a1•(4a1+6d),化简可得d=2a1∴a3a1=a1+2da1=
(I)设等差数列{an}的公差为d,由题意知d为非零常数∵a1=1,a1、a3、a9成等比数列∴a32=a1×a9,即(1+2d)2=1×(1+8d),解之得d=1(舍去0)因此,数列{an}的通项公
a2=a1+da3=a1+2da6=a1+5d由等比数列性质(a1+2d)^2=(a1+d)(a1+5d)a1=-1/2dq=a3/a2=3