已知等差数列an若a1 .. an=p an-9 ... an=q
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1.有误吧a2=a4+6=.有这个已知条件的吧2.选Ba17-a15=2d=a7-a5=-23.a1+d(n-1)=-23+4(n-1)=153n=45选B
a2005*a20060,a20050,则a2007+a2006>0因为a2005+a2006=a1+a40100所以使前n项之和sn
很简单的!a1+a5+a9=3a5=2兀.所以:a5=2兀/3.再答:又因为a2+a8=2a5.所以c0s(a2+a8)=c0s(2a5)=c0s(4兀/3)=—1/2.再答:明白了吗?
a3=a1+2d=a1+4a4=a1+3d=a1+6因为a1,a3,a4成等比数列,则a4/a3=a3/a1(a1+4)^2=a1(a1+6)解之,a1=-8则a2=a1+d=-8+2=-6
因为a1+a2+a3=15所以3a2=15a2=5a2-3=2因为a1+1,a2-3,a3-7成等比数列所以(a1+1)*(a3-7)=4设公差为x所以(5-d+1)*(5+d-7)=4所以d=4a1
已知等差数列{an}的公差d=3且a1,a3,a4成等比数列∴a3²=a1×a4∴(a2+d)²=(a2-d)×(a2+2d)∴(a2+3)²=(a2-3)×(a2+6)
a3/a1=a4/a3即为:(a1+2d)/a1=(a1+3d)/(a1+2d)因为d=2,即为a1²+,8a1+16=a1²+6a1即得a1=-8故a2=-8+2=-6
已知等差数列{an}的公差为2,若a1,a3,a4成等比数列,则a2=?a1a3a4成等比数列a3²=a1*a4(a1+2d)²=a1(a1+3d)a1²+4a1d+4d
a1+a3+a8=a1+(a1+2d)+(a1+7d)=3a1+9d=15所以a1+3d=5所以a4=a1+3d=5
(I)设数列{an}的公差为d,由已知有a1=3a1+3d=12(2分)解得d=3(4分)∴an=3+(n-1)3=3n(6分)(Ⅱ)由(I)得a2=6,a4=12,则b1=6,b2=12,(8分)设
∵首项a1=1,公差d=3,an=2014,∴2014=1+3(n-1),解得n=672.故选:C.
∵a1=13,a2+a5=4,∴2a1+5d=4,即d=23,∵an=33=a1+(n-1)d,∴13+23(n−1)=33,解得n=50,故答案为:50
A2=-6(A2+2)/(A2-2)=(A2+4)/(A2+2)
a1+...a100=0则50*(a50+a51)=0即a50+a51=0由于a10,a500,因此b1,.b48都小于0b49=a49a50a51>0b50=a50a51a520,b51以上都大于0
证明:取倒数1/an+1=an+3/3an=1/3+1/an1/an+1-1/an=1/3a1=1/21/a1=2{1/an}2首项1/3公差等差数列an=3/(5+n)
∵等差数列an中,a1+a5=2a3,又由题意a1+a3+a5=9,∴3a3=9,a3=3,则a2+a4=2a3=6,故答案为:6.
S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+
an=3n-1由an+1=an+3得知公差d=3所以an=a1+(n-1)d=3n-1
a7+a15=0a8+a14=a9+a13=a10+a12=2a11=0前11项均不大于0所以S10=S11均属于最小
设等差数列{an}的公差为d,(d>0)则1+2d=(1+d)2-4,即d2=4,解得d=2,或d=-2(舍去)故可得an=1+2(n-1)=2n-1,Sn=n(1+2n−1)2=n2,故答案为:2n