在等比数列{an}中,a1=二分之一,an=二分之二百四十三,Sn=182.求q与n,学霸们快帮帮
\u5728\u7b49\u6bd4\u6570\u5217\u4e2dSn=189,q=2\uff0can=96\uff0c\u6c42a1\u548cnan=a1*q^(n-1)
96=a1*2^(n-1)
192=a1*2^n
Sn=(a1-a1q^n)/(1-q)
189=(a1-192)/(1-2)
189=-a1+192
a1=3
192=3*2^n
64=2^n
n=6
\u89e3\uff1a
\u9996\u5148\u516c\u6bd4q\u22601\uff0c\u5426\u5219an=a1\uff0c\u77db\u76fe\uff0c
Sn=a1*(1-q^n)/(1-q)=-341
a1=1\uff0can=a1*q^(n-1)=-512
\u5373q^(n-1)=-512
\u2234[1-(-512q)]/(1-q)=-341
1+512q=341q-341
171q=-342
q=-2
\u2234q^(n-1)=(-2)^(n-1)=-512
n-1=9
n=10
\u8c22\u8c22
1/2*q^(n-1)=243/2 ,(1)
1/2*(1-q^n)/(1-q)=182 ,(2)
由(1)得 q^n=243*q ,代入(2)得 (1-243q)/(1-q)=364 ,
解得 q= 3 ,
由 3^(n-1)=243=3^5 得 n=6 。
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