www.1862.net > 已知等差数列{An}中,A1=1,公差D>0,且A2,A5,A...

已知等差数列{An}中,A1=1,公差D>0,且A2,A5,A...

∵a1、a2、a5成等比数列,∴a22=a1?a5,∴(1+d)2=1×(1+4d),d≠0.解得d=2.∴a2014=a1+2013d=1+2013×2=4027,故答案为:4027;

请采纳

∵等差数列{an}的公差和首项都不等于0,且a2,a4,a8成等比数列,∴a42=a2a8,∴(a1+3d)2=(a1+d)(a1+7d),∴d2=a1d,∵d≠0,∴d=a1,∴a1+a5+a9a2+a3=15a15a1=3.故选:B.

a2=b2 a8=b4 a32=b6 1+d=b2,......................(1) 1+7d=b1q^3=b2*q^2..............(2) 1+31d=b1q^5=b2*q^4............(3) (2)-(1)得 6d=b2(q^2-1)................................(4) (3)-(1)得 30d=b2(q^4-1)............................

数学知识都还给老师了哈哈,回答你第一问。 a1+a2=2a1+d=1 a2,a3,a5成等比数劫,a5/a3=a3/a2,即(a1+4d)/(a1+2d)=(a1+2d)/(a1+d),化简这个等式,得到a1d=0,因为d不等于0,所以a1=0。 2a1+d=1,得知d=1 所以数列{an}的通项公式为:an=a1+(n-1...

格式o(╯□╰)o,看不清

行列式的所有元素的代数余子式为.a5a6a8a9.,-.a4a6a7a9.,.a4a5a7a8.,-.a2a3a8a9.,.a1a3a7a9.,-.a1a2a7a8.,.a2a3a5a6.,-.a1a3a4a6.,.a1a2a4a5.,由于不考虑符号,各行列式的值均小于0,比如.a5a6a8a9.=(a1+4d)(a1+8d)-(a1+5d)(a1+...

∵数列{an}的奇数项成等差数列,偶数项成等比数列,公差与公比均为2,∴a3=a1+2,a5=a1+4,a7=a1+6,a4=2a2,a6=4a2,∵a2+a4=a1+a5,a4+a7=a6+a3∴a2+2a2=a1+4+a1,2a2+6+a1=4a2+2+a1∴a1=1,a2=2,∵am?am+1?am+2=am+am+1+am+2成立,∴由上面可以知数...

请采纳

(1)∵a5和a7的等差中项为11,∴a6=11,又a2?a5=a1?a14.可得a1+5d=11(a1+d)(a1+4d)=a1(a1+13d),又公差d≠0,解得a1=1d=2∴an=a1+(n-1)d=1+2(n-1)=2n-1.∴bn=1(2n?1)(2n+1)=12(12n?1?12n+1).∴Tn=12[(1?13)+(13?15)+…+(12n?1?12n+1)]=12(...

网站地图

All rights reserved Powered by www.1862.net

copyright ©right 2010-2021。
www.1862.net内容来自网络,如有侵犯请联系客服。zhit325@qq.com