已知等差数列an的公差d 4a2 a5 a7最小值
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即对任意n∈N,(a+n)/(a+n-1)≥(a+8)/(a+7)两边同减1:1/(a+n-1)≥1/(a+7)此不等式可分三种情况:(1)a+7≥a+n-1〉0显然n≥8时不成立(2)0〉a+n-1
∵an为等差数列a1,a3,a9成等比数列∴a1(a1+8d)=(a1+2d)^2a1^2+8d*a1=a1^2+4d*a1+4d^2d≠0∴d=a1a1+a3+a9/a2+a4+a10=(a1+a1
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
由题意可得an=a1+(n-1)d=1+23(n-1)=2n+13,∵bn=(-1)n-1anan+1,∴当n为偶数时,Sn=b1+b2+…+bn=a1a2-a2a3+a3a4-a4a5+…+an-1
bn=sn-s(n-1)=1-1/3^n-(1-1/3^n-1)=-1/3^n+3/3^n=2/3^n
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
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
(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
a2+a4=2*a3=8a3=4,a4=3因此a1=6,d=-1通项为an=6-(n-1)=7-n
1.S5=5a1+10d=5(a1+2d)=70a1+2d=14a3=14a7^2=a2×a22(a3+4d)^2=(a3-d)(a3+19d)a3=14代入,整理,得d(d-4)=0d=0(已知d不
6m+7=3k+16(m+1)=3kk=2m+2q=bn/bn-1=an+1/an-1an+1-(an-1)=2d两个联立an-1=1+2d/q是常数所以an是常数列bn也是常数列,且bn=1
因为a1+a5=a2+a4=4,所以:a2a4=3a2+a4=4解方程组:a2=1a4=3或者a2=3a4=1a4-a2=2d=2,或者a4-a2=-2d=1,或者d=-1
ak=48+2kbk=10+(k-1)dSk=(48+2k)[10+(k-1)d]令SK≤21即(48+2k)[10+(k-1)d]≤21求出k来.再问:最大圆面积为Sk
a1,a5,a17是等比数列(a1+4d)^2=a1*(a1+16d)a1^2+8a1d+16d^2=a1^2+16a1d8a1d=16d^2d不等于0a1=2dq=a5/a1=(a1+4d)/a1=
(1).由a(m)+a(m+1)=a(k)知道3m+3(m+1)+1=3k+1,整理后有k-2m=4/3,而m,k均是N+,则k-2m也是整数,故而不存在m,k∈N+,使a(m)+a(m+1)=a(k
因为{An}是等差数列,所以A2+A8=A4+A6=10,A4*A6=24,所以可将A4、A6看作方程x^2-24x+10=0的两个根,因为d
先做个mark,回头再做给你看.----------------------------------------将{an}分拆成{bt}、{ct}数列排列如下:{bt}:a1,a3,a5,a7,a9,
5或6是对的,a6=0,S5=S6,a1^2=a11^2a11^2-a1^2=0(a11+a1)(a11-a1)=0(2a1+10d)*10d=0d
再问:太给力了,你的回答完美解决了我的问题!
先求An的通项就行了A1+A4=14A2A3=45d