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等比数列A2=2,A2+A4+A6=14,A6=?

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解: S100=a1+a2+...+a100 =(a1+a3+...+a99)+(a2+a4+...+a100) =(a2/q+a4/q+...+a100/q)+(a2+a4+...+a100) =(1/q)(a2+a4+...+a100)+(a2+a4+...+a100) =(1+ 1/q)(a2+a4+...+a100) a2+a4+...+a100=S100/(1+ 1/q)=150/[1+ 1/(1/2)]=50

解:设公比为q,an>0,则q>0 a2a4+2a3a5+a4a6=36 a32+2a3·a5+a52=36 (等比中项性质) (a3+a5)2=36 an>0,a3>0,a5>0,a3+a5>0 a3+a5=6

解: 设公比为q,an>0,则q>0 a2a4+2a3a5+a4a6=36 a3²+2a3·a5+a5²=36 (等比中项性质) (a3+a5)²=36 an>0,a3>0,a5>0,a3+a5>0 a3+a5=6

∵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...

(Ⅰ)∵{an}是正项等比数列,a2a8=4,∴a52=4,解得a5=2,又∵a4+a6=203,∴a1q4=2a1q3+a1q5=203,两式相除得:q1+q2=310.…(2分)解得q=3或者q=13,∵{an}为增数列,∴q=3,a1=281.…(4分)∴an=a1qn-1=281?3n-1=2?3n-5.∴bn=log3an2=n-5.…(6...

解: a7=a2+5d d=(a7-a2)/5=(7-22)/5=-3 a20=a2+18d=22+18·(-3)=-32 a(n+2)-an=2d=2·(-3)=-6 数列{an}是偶数项是以22为首项,-6为公差的等差数列 20/2=10,数列共10项。 a2+a4+a6+...+a20 =(a2+a20)·10/2 =(22-32)·10/2 =-50

由:第2,3项及第6项构成等比数列,可列:(a1+d)(a1+5d)=(a1+2d)^2因公差不为0化简得:d=-2a1 (a1+a3+a5)/(a2+a4+a6)=(3a1+6d)/(3a1+9d)将d=-2a1代入得: (a1+a3+a5)/(a2+a4+a6)=(3a1+6d)/(3a1+9d)=3/5 谢谢加分!!

a2+2a6+a10=(a2+a6)+(a6+a10) =(a4+a8)/q^2+(a4+a8)*q^2 =(a4+a8)(1/q^2+q^2)=-2(1/q^2+q^2)................(1) 又知a4+a8=a6(1/q^2+q^2)=-2.................................(2) 结合(1)(2)得 a6(a2+2a6+a10)=-2a6(1/q^2+q^2)=-2*(-2)=4 ...

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