已知数列an为等差数列,且a1=50d=-0.6

已知数列{log2(an-1)}为等差数列,且a1=3,a2=5可以得到该等差数列的公差d：d＝log2(a2-1)-log2(a1-1)＝log2(5-1)-log2(3-1)＝log2(4)-lo

a2005*a20060,a20050,则a2007+a2006>0因为a2005+a2006=a1+a40100所以使前n项之和sn

1+Sn=2anSn=2an-1n>=2则S(n-1)=2a(n-1)-1相减an=2an-2a(n-1)an=2a(n-1)所以n>=2时是等比q=2a1=S1所以1+a1=2a1a1=1所以an=

①设公差为d，公比为q∵数列{an+bn}的前三项依次为3，7，13∴a1+b1＝3a2+b2＝7a3+b3＝13又a1=1∴b1＝2d＝2q＝2∴an=2n-1，bn=2n②∵an=2n-1，bn=

a(n)=2n-1b1=12b2=b1公比为1/2b(n)=1/2^(n-1)Cn=(2n-1)*2^(n-1)Sn=1+3*2^1+5*2^2+.+(2n-1)*2^(n-1)2Sn=2+3*2^2

(1)设bn=log2(an+1),则{bn}为等差数列,又a1=1,a3=7,所以b1=log2(1+1)=1,b2=log(7+1)=3,所以公差d=1.所以bn=b1+(n-1)d=1+(n-1

设公差值为ca1+a2+a3=a1+（a1+c）+(a1+c+c)=3a1+3c=12c=2an=a1+c(n-1)=2nbn=3^(2n)b(n+1)/bn=3^(2n+2)/3^2n=9所以bn是

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a1*p=a2a1*p^3=a4,a1*p-a1=a1*p^3-a1*Pp-1=p^(p^2-1);(p-1)(p*(p+1)-1)=0,p=1,或p^2+p-1=0,p=(-1+√5)/2,p=(-

a1,a2,a4成等差数列2a2=a1+a4即2a1*q=a1+a1q^3a1不为0所以：2q=1+q^3q^3-2q+1=0q^3-q^2+q^2-2q+1=0q^2*(q-1)+(q-1)^2=0

a1,a2,a4成等差数列所以2a2=a1+a4{an}是等比数列a2=a1qa4=a1q^3所以2×a1q=a1+a1q^3即：q^3-2q+1=0(q-1)(q^2+q-1)=0q=1或q=(-1

∵1,an,Sn为等差数列∴2a1=1+S1=1+a12a2=1+S2=1+a1+a2∴a1=1a2=2由2an=1+Sn2a(n-1)=1+S(n-1)得2an-2a(n-1)=Sn-S(n-1)=

设公差为d,a1=a2-d,a3=a2+d,则原等式变为,a2-d+a2+a2+d=15消去d,3a2=15a2=5

1.a1=1,a2=3,所以an=2n-1b1=1,b2=0.5,所以an=(0.5)^(n-1)=2^(1-n)2.Cn=an/bn=(2n-1)*2^(n-1)Sn=1*2^0+3*2^1+5*2

a1=a2-2，a5=a2+6∴a22=a1a5=（a2-2）（a2+6），解得a2=3故选D

设数列log2(an-1)公差为dd=long2(an-1)-log2(a(n-1)-1)=log2[(an-1)/(a(n-1)-1]所以(an-1)/(a(n-1)-1)=2^d而由a1=3a2=

设首项为a1,公差为d.由题得：a1+a5=2*4a1*a7=a₃^2则:a1+(a1+4d)=8a1(a1+6d)=(a1+2d)^2综上解得a1=2d=1所以S5=20

设an=a1+(n-1)d=10+(n-1)dSn=na1+(n-1)nd/2=10n+(n-1)nd/2S12=120+66d=-125那么d就算出来了d=-245/66所以an=10+(n-1)(

∵数列{an}为等差数列，∴a1+a13=a2+a12=2a7，∵a1+a7+a13=π，∴3a7=π，解得a7＝π3．则tan（a2+a12）=tan(2a7)＝tan2π3=-tanπ3＝−3．故