题目
Problem
(a) The first three terms of an arithmetic progression are 25, 4p−1 and 13−p, where p is a constant.
Find the value of the tenth term of the progression.
[4]
(b) The first three terms of a geometric progression are 25, 4q−1 and 13−q, where q is a positive constant.
Find the sum to infinity of the progression.
[4]
题目中文翻译
(a) 一个等差数列的前三项为 25、4p−1 和 13−p，其中 p 为常数。
求该数列的第十项。
[4]
(b) 一个等比数列的前三项为 25、4q−1 和 13−q，其中 q 为正常数。
求该数列的无穷和。
[4]
解答
(a)
解法一
思路
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等差中项等于两端的平均。先求 p，再求公差和第十项。
答题过程
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The arithmetic condition is 2(4p−1)=25+(13−p). Thus 9p=40, so p=40/9. The common difference is
(4p−1)−25=4(40/9)−26=−74/9.
The tenth term is 25+9(−74/9)=−49.
(b)
解法一
思路
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等比中项的平方等于两端的乘积。先求正的 q，再检查公比能否用来求无穷和。
答题过程
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We have
(4q−1)2=⟹16q2+17q−324=25(13−q)(q−4)(16q+81)=0.
Since q>0, q=4. Then the ratio is (16−1)/25=3/5, so the sum to infinity is
1−3/525=2125.
解法二
思路
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也可以直接设公比，用首项 25 表示第二项和第三项，然后消去 q。
答题过程
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With common ratio r, the conditions are 25r=4q−1 and 25r2=13−q. Substitution of q=(25r+1)/4 gives
100r2+25r−51=(5r−3)(20r+17)=0.
The root r=−17/20 gives q<0, so only r=3/5 is allowed. Thus the sum to infinity is 125/2.
