已知an>0,且a1,1 2a3,a2成等差数列,求a3 a4 a4 a5.
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等差数列{an}中,a1=a1,a3=a1+2d,a9=a1+8d,因为a1、a3、a9恰好是某等比数列,所以有a32=a1a9,即(a1+2d)2=a1(a1+8d),解得d=a1,所以该等差数列的
∵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
等差数列a1+a2+a3=3*a2=12a2=4d=(a8-a2)/6=2a1=a2-d=2所以an=2+(n-1)*2=2n
a3=a1+2dq9=a1+8da1,a3,a9成等比数列所以(a1+2d)^2=a1*(a1+8d)a1^2+4a1d+4d^2=a1^2+8a1dd^2=a1dd≠0d=a1所以a1+a3+a9=
anan+1-2an=0anan+1=2anan+1=2所以a2=2a3=2a4=2
a3²=a1a13(a1+2d)²+a1(a1+12d)a1=1所以1+4d+4d²=1+12d4d²-8d=0所以d=2所以an=2n-1bn=2^)2n-1
a2=a1+d,a3=a1+2d.,a6=a1+5d,...,a10=a1+9d,若a1,a3,a6成等比数列,则a3^2=a1*a6,(a1+2d)^2=a1*(a1+5d),得到a1=4d.则(a
a1=2a1+a2+a3=12a2=4d=2an=2nbn=3^an=3^2n=9^n数列bn是以9为首项,公比=9的等比数列Sn=9(1-9^n)/(1-9)=(9^[n+1]-9)/8
a1=2,a1+a2+a3=12a2=4d=2an=2n2.Sn=2*3+4*3^2+6*3^3+……+2n*3^n3Sn=2*3^2+4*3^3+……+(2n-2)*3^n+2n*3^[n+1]相减
{an}是等差数列,且a1=2,a1+a2+a3=12而2a2=a1+a3所以a2=4所以公差d=a2-a1=2所以an=a1+(n-1)d=2nbn=(1/2)^n*2n和Tn=b1+b2+……+b
(1)由题设可知公差d≠0,由a1=1且a1,a3,a9成等比数列,得:(1+2d)2=1+8d,解得d=1或d=0(舍去),故{an}的通项an=n.(2)∵bn=2 an=2n,∴数列{
a1,a3,a9成等比数列a3^2=a1*a9(a1+2d)^2=a1*(a1+8d)解得a1=d(a1+a3+a9)/(a2+a4+a10)=(3a1+10d)/(3a1+13d)=13d/16d=
设an的公差是d∴a3=a1+2d,a9=a1+8da2=a1+d,a4=a1+3d,a10=a1+9d∴a1+a3+a9=3a1+10d,a2+a4+a10=3a1+13d∵a1,a3,a9依次成等
最后应该是a10吧a1*a9=(a3)²a1*(a1+8d)=(a1+2d)(a1+2d)化简得a1=d(a1+a3+a9)/(a2+a4+a10)=(a1+a1+2d+a1+8d)/(a1
a1=a3-2d,a9=a3+6d因为a1,a3,a9成等比数列,所以有(a3)^2=(a1)*(a9)所以(a3)^2=(a3-2d)(a3+6d)所以3d^2=d*(a3)因为d不等于0所以a3=
由等差数列的性质可得2a2=a1+a3=8,解得a2=4,又a2+a4=12,所以a4=12-4=8,故数列的公差d=a4−a22=2,故an=a2+(n-2)d=4+2(n-2)=2n,故答案为:2
An=6Sn/(An+3)6Sn=(An)^2+3Ann>=26S(n-1)=(A(n-1))^2+3A(n-1)6An=(An)^2+3An-(A(n-1))^2-3A(n-1)(An)^2-(A(
【1】a1a2a3=a8-7da8-6da8-5d=3a8-18d=48-18d=12d=2a1=a8-7d=2an=a1(n-1)d=2n【2】bn=2n*2=4n再问:第二问怎么出来的
由题可得A1*A9等于A3方把分子分母都写为A3和公差d的表达式有上式可得A3和d的关系带入就可的到比值
∵an+1=an+cn∴an+1-an=cn∴an-an-1=c(n-1)an-1-an-2=c(n-2)…a2-a1=c×1上述各式相加得:an-a1=cn(n-1)/2∴a2-a1=ca3-a1=