已知an为等差数列_a1等于2
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![已知an为等差数列_a1等于2](/uploads/image/f/4216334-14-4.jpg?t=%E5%B7%B2%E7%9F%A5an%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97_a1%E7%AD%89%E4%BA%8E2)
由{1/(an+1)}为等差数列,所以1/(a3+1)=1/3,1/(a7+1)=1/2,1/(a11+1)成等差数列,从而1/(a11+1)=2/3,解得a11=1/2
(1)∵a[1]=2,a[n]=2a[n-1]-1,n≥2∴a[n]-1=2(a[n-1]-1)n≥2∵a[1]-1=1∴{a[n]-1}是首项为1公比为2的等比数列∵a[n]-1=2^(n-1)∴a
http://zhidao.baidu.com/question/88231937.html?fr=qrl&cid=983&index=2S1=a1=-(2/3),S2+1/S2+2=a2,因为S2=
奇数项有n+1偶数项是n则奇数和=[a1+a(2n+1)](n+1)/2=290偶数和=[a2+a(2n)]n/2=261等差则a1+a(2n+1)=a2+a2n所以相除有(n+1)/n=290/26
S1=a1=-(2/3),S2+1/S2+2=a2,因为S2=(a1+a2),所以S2+1/S2+2=S2-a1=S2+2/3,解得S2=-(3/4),同理,S3+1/S3+2=a3=S3-S2=S3
已知等差数列{an}的前三项依次为a-1,a+1,2a+3,故有2(a+1)=a-1+2a+3,解得a=0,故等差数列{an}的前三项依次为-1,1,3,故数列是以-1为首项,以2为公差的等差数列,故
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
a(n)=a(n+3).不可能递增.
再问:你做错了喔再问:答案等于145
因为1/(an+1)为等差数列,设bn=1/(an+1)b3=1/3b7=1/2,b7=b3+4dd=(b7-b3)/4公差d=(1/2-1/3)/4=1/24b8=b7+d=1/2+1/24=12/
因为等差数列{an}的公差为-2,且a2,a4,a5成等比数列,所以a42=a2a5,即(a2−4)2=a2(a2−6).整理得,2a2=16,所以a2=8.故答案为8.
a3=2,a7=1,若{1/(an+1)}为等差数列公差d=(1/a7+1-1/a3+1)/(7-3)=1/241/(a11+1)=1/3+1/24*(11-3)=2/3a11=1/2
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
因为是等差的,所以和的个数是偶数的话,和=中间两项相加*个数/2也就是说=(a4+a5)*8/2=72(8就是一共有8个数相加,a4、a5为中间两项)如果和的个数是奇数的话,和=中间一项*2*(个数+
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)
Sn与2的等比中项为√(2Sn),an与2的等差中项为(an+2)/2由题目可知,8Sn=(an+2)^2,所以8S_(n-1)=[a_(n-1)+2]^2.两者相减,得8an=an^2+4an-[a
an=2n-3x+1-(x-1)=2x+3-(x+1)x=0d=2an=-1+(n-1)x2=2n-3
a2=a1+da5=a1+4d2a1+5d=4==>d=2/3an=a1+(n-1)d=331+2(n-1)=99n=50
由题意可得,a32=a1a4∴(a1+4)2=a1(a1+6)∴a1=-8∴S9=9×(−8)+9×8×22=0故选A