設(shè){an}為等差數(shù)列,{bn}為等比數(shù)列,a1=b1=1,a2+a4=b3,b2b4=a3,分別求出{an}及{bn}的前10項的和S10及T10.
【答案】
分析:根據(jù)等差數(shù)列的性質(zhì)可知a
2+a
4=2a
3,根據(jù)等比數(shù)列的性質(zhì)可知b
2b
4=b
32,而已知a
2+a
4=b
3,b
2b
4=a
3,所以得到b
3=2a
3,a
3=b
32,兩者聯(lián)立,由b
3≠0,即可求出a
3與b
3的值,然后分別根據(jù)a
1=b
1=1,利用等差及等比數(shù)列的通項公式求出等差數(shù)列的公差d及等比數(shù)列的公比q,然后根據(jù)等差、等比數(shù)列的前n項和的公式即可求出{a
n}及{b
n}的前10項的和S
10及T
10的值.
解答:解:∵{a
n}為等差數(shù)列,{b
n}為等比數(shù)列,
∴a
2+a
4=2a
3,b
2b
4=b
32已知a
2+a
4=b
3,b
2b
4=a
3,
∴b
3=2a
3,a
3=b
32得b
3=2b
32∵b
3≠0∴
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由a
1=1,
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知{a
n}的公差為
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,
∴
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,
由b
1=1,
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知{b
n}的公比為
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或
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.
當(dāng)
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時,
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,
當(dāng)
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時,
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.
點評:此題考查學(xué)生靈活運用等差、等比數(shù)列的通項公式及等差、等比數(shù)列的前n項和的公式化簡求值,是一道綜合題.