www.1862.net > 已知等比数列{An}中A(n+1)>An,且A3+A7=3,A2A8...

已知等比数列{An}中A(n+1)>An,且A3+A7=3,A2A8...

(1)q^3=a7/a4=8/2=4 q=4^(1/3)=2^(2/3) a1=a4/q^3=2/4=1/2 an=a1*q^(n-1)=1/2*(2^(2/3))^(n-1)=2^(-1+2n/3-2/3)=2^(2n/3-5/3) (2) a2+a5=a2(1+q^3)=18 a3+a6=a3(1+q^3)=9 下式/上式得:q=a3/a2=1/2 a2+a5=a1*1/2+a1*(1/2)^4=18 a1=32 an=a1*q...

∵数列{an}是各项均为正的等比数列,∴数列{an+an+1}也是各项均为正的等比数列,设数列{an+an+1}的公比为x,a1+a2=a,则x∈(1,+∞),a3+a4=ax,∴有a3+a4-a2-a1=ax-a=8,即a=8x?1∴y=a5+a6+a7+a8=ax2+ax3=8(x3+x2)x?1,x∈(1,+∞),求导数可得y′=1...

∵等比数列{an}满足a1?a4?a7=1,a2?a5?a8=8,∴a43=1,a53=8,∴a4 =1,a5 =2.故公比为q=a5a4=2.∴a3?a6?a7=(a2?a5?a8 )q=8×2=16,故选C.

an=a1.q^(n-1) a2.a3.a4=8 (a1)^3.q^6=8 (1) a6.a7.a8=64 (a1)^3.q^18=64 (2) (2)^2/(1) (a1)^3.q^30 =512 a10.a11.a12 =(a1)^3.q^30 =512

利用等比数列{an}的性质有S2,S4-S2,S6-S4,S8-S6成等比数列,∴S2=40,S4-S2=a3+a4=60,则S6-S4=90,S8-S6=135故a7+a8=S8-S6=135.故选A

a1+a2+a3/a7+a8+a9=1/q^6 q^6=4 之后可得a1 可得嗄 a10+a11+a12+…+a3m+1+a3m+2+a3m+3+(a4+a5+...+a9)=所以 a4(1-q^m)/1-q 之后得s后减去(s9-s3)即可,如果有不明白的发上来我再告诉你

∵等差数列{an}的前n项和为Sn,且a2=-9,a3+a7=-6,∴a1+d=?9a1+2d+a1+6d=?6,解得a1=-11,d=2,∴Sn=-11n+n(n?1)2×2=n2-12n=(n-6)2-36≥-36,∴当且仅当n=6时,Sn取最小值-36.故选:D.

∵等比数列{an}满足a1?a4?a7=1,a2?a5?a8=8∴a43=1即a4=1,a53=8即a5=2∴公比q=a5a4=2,则a6=4a3?a6?a9=a63=43=64故选D.

∵等差数列{an}中,a3+a7-a10=8,a11-a4=4,∴(a3+a7-a10)+(a11-a4)=(a3+a11)-(a4+a10)+a7=8+4=12,由等差数列的性质可得a3+a11=a4+a10,∴a7=12,∴S13=13(a1+a13)2=13×2a72=13a7=13×12=156故选:C.

∵数列{an}为等差数列,∴a1+a2+a3,a4+a5+a6,a7+a8+a9构成等差数列,∴a7+a8+a9=9+2×18=45,故选B

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