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CIE 9709 2025 Nov Paper 15 Q2

A Level / CIE / P1

CIE 9709 2025 Nov Paper 15 Paper · Question 2

题目

Problem

A geometric progression has first term 3+423+4\sqrt{2} and second term 5−25-\sqrt{2}.

(a) Find the common ratio of the geometric progression. Give your answer in the form 2+p\sqrt{2}+p, where pp is an integer to be found.

[3]

(b) Find the sum to infinity of the geometric progression.

[2]
题目中文翻译

一个等比数列的第一项为 3+423+4\sqrt{2},第二项为 5−25-\sqrt{2}。

(a) 求这个等比数列的公比。答案写成 2+p\sqrt{2}+p 的形式,其中 pp 是需要求出的整数。

(b) 求这个等比数列的无穷和。

解答

(a)

解法一

思路

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第二项除以第一项得到公比。用分母的共轭式有理化,便能读出整数项。

答题过程

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The common ratio is

r= 5−23+42⋅3−423−42= 15+8−2329−32=2−1.\begin{align*} r=&\,\frac{5-\sqrt2}{3+4\sqrt2}\cdot\frac{3-4\sqrt2}{3-4\sqrt2}\\ =&\,\frac{15+8-23\sqrt2}{9-32} =\sqrt2-1. \end{align*}

Hence p=−1p=-1.

解法二

思路

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题目已经给出公比的形式。用“第一项乘公比等于第二项”建立等式,比较有理部分即可求出整数 pp。

答题过程

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Since the common ratio has the form 2+p\sqrt2+p with pp an integer,

(3+42)(2+p)=(8+3p)+(3+4p)2=5−2.(3+4\sqrt2)(\sqrt2+p) =(8+3p)+(3+4p)\sqrt2=5-\sqrt2.

Comparing rational parts gives 8+3p=58+3p=5, so p=−1p=-1. This also gives 3+4p=−13+4p=-1, as required for the coefficient of 2\sqrt2. Hence the common ratio is 2−1\sqrt2-1.

(b)

解法一

思路

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先检查公比的绝对值小于 11,再用等比数列的无穷和公式。

答题过程

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Since 0<2−1<10<\sqrt2-1<1, the sum to infinity exists and is

S∞= 3+421−(2−1)=(3+42)(2+2)(2−2)(2+2)= 14+1122=7+1122.\begin{align*} S_\infty =&\,\frac{3+4\sqrt2}{1-(\sqrt2-1)} =\frac{(3+4\sqrt2)(2+\sqrt2)}{(2-\sqrt2)(2+\sqrt2)}\\ =&\,\frac{14+11\sqrt2}{2} =7+\frac{11\sqrt2}{2}. \end{align*}