在等比数列{an}中,(1)已知a1=9⼀8,an=1⼀3,q=2⼀3,求n; (2)已知:a3+a6=36,a4+a7=18,an=1⼀2,求n.

2025-01-07 05:21:37
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回答1:

(1)an=a1q^(n-1)
1/3=9/8*(2/3)^(n-1)
8/27=(2/3)^3=(2/3)^(n-1)
n-1=3,n=4
(2) a3+a6=36,---->a3(1+q^3)=36(1)
a4+a7=18,----->a4(1+q^3)=18(2)
(2)/(1),得
a4/a3=1/2=q
代入(1)式,得
a3=36/(1+(1/2)^3)=32
an=a1q^(n-1)=a3q^(n-3)
1/2=32*(1/2)^(n-3)
1/(2*32)=1/2^6=1/2^(n-3)
n-3=6,--->n=9