已知等差数列{an}的前n项和为Sn,公差d≠0,且S3+S5=58,a1,a3,a7成等比数列.(I)求数列{an}的通项公
已知等差数列{an}的前n项和为Sn,公差d≠0,且S3+S5=58,a1,a3,a7成等比数列.(I)求数列{an}的通项公式;(II)若{bn}为等比数列,且b5?b...
已知等差数列{an}的前n项和为Sn,公差d≠0,且S3+S5=58,a1,a3,a7成等比数列.(I)求数列{an}的通项公式;(II)若{bn}为等比数列,且b5?b6+b4?b7=a8,记Tn=log3b1+log3b2+…+log3bn,求T10值.
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(Ⅰ)由S3+S5=58,得3a1+3d+5a1+10d=8a1+13d=58,①
∵a1,a3,a7成等比数列,a32=a1a7,
即(a1+2d)2=a1(a1+6d),整理得a1=2d,
代入①得d=2,a1=4,
∴an=2n+2. …(6分)
(Ⅱ)由(Ⅰ)知a8=18,b5?b6+b4?b7=2b5?b6=18,解得b5?b6=9.
∵T10=log3b1+log3b2+log3b3+…+log3b10
=log3(b1?b10)+log3(b2?b9)+…+log3(b5?b6)
=5log3(b5?b6)=5log39=10. …(12分)
∵a1,a3,a7成等比数列,a32=a1a7,
即(a1+2d)2=a1(a1+6d),整理得a1=2d,
代入①得d=2,a1=4,
∴an=2n+2. …(6分)
(Ⅱ)由(Ⅰ)知a8=18,b5?b6+b4?b7=2b5?b6=18,解得b5?b6=9.
∵T10=log3b1+log3b2+log3b3+…+log3b10
=log3(b1?b10)+log3(b2?b9)+…+log3(b5?b6)
=5log3(b5?b6)=5log39=10. …(12分)
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