在公差不为零的等差数列a3=7
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数列a1a2a5等比数列则有a2*a2=a1*a5a3-2d=a1a3+2d=a5a3-d=a2带入得到d=2b1+2b2+4b3+2^(n-1)bn=an(1)b1+2b2+4b3+2(n-3)bn
(1)∵等差数列{an}中,a2,a4,a9成等比数列,∴a42=a2•a9,即(a1+3d)2=(a1+d)(a1+8d),整理得:6a1d+9d2=9a1d+8d2,即d2=3a1d,∵d≠0,∴
(Ⅰ)设公差为d,由条件得5a1+5×42d=30(a1+2d)2=a1(a1+8d),得a1=d=2.∴an=2n,Sn=2n+n(n-1)×22=n2+n;(Ⅱ)∵1Sn+an+2=1n2+n+2
(1)a3=a1+2d,a7=a1+6d,所以a1*a7=a3*a3,即a1*(a1+6d)=(a1+2d)*(a1+2d)解得d=1(2)Sn=(1/2)n^2+(3/2)n,又a3=a1+2d=4
a1+q^2*a1=2*q*a1解得q=1不存在满足条件的答案……你检查题目是不是有问题……
因为b1=a1²,b2=a2²;所以b1>0,q>0且q≠1({an}公差不为零)所以a1=√b1,a2=√(b1*q),a3=q*√b12a2=a1+a3->2√(b1*q)=√
令{an}公差为d,由b2^2=b1*b3得:a2^4=a1^2*a3^2两边开方得:a2^2=a1*a3或a2^2=-a1*a3当a2^2=a1*a3时,有: (a
(1)由题意,设公差为d,则a1+4d=10(a1+2d)2=a1(a1+8d)∴a1+4d=104d2=4a1d∵d≠0,∴a1=2,d=2∴an=2+(n-1)×2=2n;(2)由(1)知,Sn=
a2^2+a3^2=a4^2+a5^2a2^2+(a2+d)^2=(a2+2d)^2+(a2+3d)^2解得d=2a2/3Sn7=7a1+3d=1解得d=2/7a1=1/7an=1/7+(n-1)2/
1)因为an为等差数列所以a1=5-2da2=5-da5=5+2d又a1,a2,a5成等比数列所以(a2)^2=a1*a5既(5-d)^2=(5-2d)*(5+2d)又d≠0解得d=2则a1=1an=
an等差,则a3+a11=2a72a3-a7的平方+2a11=0→4a7-a7的平方=0bn等比,则bn不为零,即a7=b7,即a7不为零所以a7=4=b7若是求b5b8,则条件不足若是求b6b8,则
(1)a3=a1+2d、a6=a1+5d.(a1+2d)^2=a1(a1+5d)a1^2+4a1d+4d^2=a1^2+5a1d4a1d+4d^2=5a1d因为d0,所以4a1+4d=5a1a1=4d
由a1+a3=4知a1+(a1+2d)=4即a1+d=2,又a2,a3,a5成等比数列得到a32=a2a5即(a1+2d)2=(a1+d)(a1+4d),a12+4da1+4d2=a12+5da1+4
(1)设等差数列{an}的公差为d(d≠0),由a1,a3,a13成等比数列,得a32=a1•a13,即(1+2d)2=1+12d得d=2或d=0(舍去).故d=2,所以an=2n-1(2)∵bn=2
设an=a+d*(n-1)1.a3+a10=a+2d+a+9d=2a+11d=152.a3*a7=a4*a4(a+2d)(a+6d)=(a+3d)^2a=-1.5d联立1与2,求得d=15/8a=-4
(1)设数列的公差为d,则∵a3=7,又a2,a4,a9成等比数列.∴(7+d)2=(7-d)(7+6d)∴d2=3d∵d≠0∴d=3∴an=7+(n-3)×3=3n-2即an=3n-2;(2)∵bn
(I)设等差数列{an}的公差为d,由题意知d为非零常数∵a1=1,a1、a3、a9成等比数列∴a32=a1×a9,即(1+2d)2=1×(1+8d),解之得d=1(舍去0)因此,数列{an}的通项公
a2=a1+da3=a1+2da7=a1+6da3^2=a2*a7(a1+2d)^2=(a1+d)(a1+6d)a1^2+4a1d+4d^2=a1^2+7a1d+6d^23a1d+2d^2=0d≠0∴
a2=a1+da3=a1+2da6=a1+5d由等比数列性质(a1+2d)^2=(a1+d)(a1+5d)a1=-1/2dq=a3/a2=3
a2,a3,a6组成等比数列的连续三项∴a3的平方=a2a6(a1+2d)²=(a1+d)(a1+5d)化简得d=-2a1q=a3/a2=(a1+2d)/(a1+d)=(-3a1)/(-a1