已知等差数列{an}中,a1+a2+a3=27,a6+a8+a10=63 (1)求数列{an}的通项公式; (2)令bn=3an,求数列{bn}
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已知等差数列{an}中,a1+a2+a3=27,a6+a8+a10=63 (1)求数列{an}的通项公式; (2)令bn=3an,求数列{bn}
已知等差数列{an}中,a1+a2+a3=27,a6+a8+a10=63 (1)求数列{an}的通项公式; (2)令bn=3an,求数列{bn}
已知等差数列{an}中,a1+a2+a3=27,a6+a8+a10=63 (1)求数列{an}的通项公式; (2)令bn=3an,求数列{bn}
(1)等差数列{an}中,由a1+a2+a3=27,a6+a8+a10=63,利用等差数列的通项公式,列出方程组,求出首项和公差,由此能求出an=2n+5.
(2)由an=2n+5,bn=3an,知bn=32n+5,由此得到数列{bn}是首项为37,公比为9的等比数列,从而能求出数列{bn}的前n项
(1)∵等差数列{an}中,a1+a2+a3=27,a6+a8+a10=63,
∴
a1+a1+d+a1+2d=27a1+5d+a1+7d+a1+9d=63
解得a1=7,d=2,
∴an=7+(n-1)×2=2n+5.
(2)∵an=2n+5 ,bn=3an,
∴bn=32n+5,
b1=37,
bn+1bn
=
32n+732n+5
=9,
∴数列{bn}是首项为37,公比为9的等比数列,
.
a6+a8+a10=63
a1+5d+a2+6d+a3+7d=63
a1+a2+a3+18d=63
27+18d=63
18d=36
d=2
a1+a2+a3=27
a1+a1+d+a1+2d=27
3a1+3d=27
3a1+3*2=27
3a1=21
a1=7
an=a1+(n-1)d
=7+2(n-1)
=2n+5
bn=3an
=3*(2n+5)
=6n+15