题目
Problem
The geometric progression a1,a2,a3,… has first term 2 and common ratio r where r>0. It is given that 29a5+7a3=8.
(a) Find the value of r.
[3]
(b) Find the sum of the first 20 terms of the geometric progression. Give your answer correct to 4 significant figures.
[2]
(c) Find the sum to infinity of the progression a2,a5,a8,….
[3]
题目中文翻译
等比数列 a1,a2,a3,… 的首项为 2,公比为 r,其中 r>0。已知 29a5+7a3=8。
(a) 求 r 的值。
[3]
(b) 求该等比数列前 20 项的和。答案保留 4 位有效数字。
[2]
(c) 求数列 a2,a5,a8,… 的无穷和。
[3]
解答
(a)
解法一
思路
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用 an=2rn−1 写第三项和第五项。得到关于 r2 的二次方程后,只保留正的公比。
答题过程
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The condition gives 9r4+14r2−8=0. Putting u=r2 gives
9u2+14u−8=(9u−4)(u+2)=0.
As u>0, u=4/9. The positive ratio is r=2/3.
(b)
解法一
思路
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用等比数列有限和公式,并保留精确公比到最后。
答题过程
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S20==1−2/32(1−(2/3)20)6(1−(2/3)20)≈5.998
to four significant figures.
(c)
解法一
思路
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新数列从原第二项开始,每次跨过三项的位置差,所以新公比是 r3。
答题过程
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The first term is a2=4/3, and the common ratio is (2/3)3=8/27. It lies between 0 and 1, so the sum to infinity is
1−8/274/3=1936.