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CIE 9709 2023 November Paper 13 Q5

A Level / CIE / P1

CIE 9709 2023 Nov Paper 13 Paper · Question 5

题目

Problem

The first, second and third terms of a geometric progression are 2p+62p+6, 5p5p and 8p+28p+2 respectively.

(a) Find the possible values of the constant pp.

(3)

(b) One of the values of pp found in (a) is a negative fraction.

Use this value of pp to find the sum to infinity of this progression.

(4)
题目中文翻译

一个等比数列的第一、第二和第三项分别为 2p+62p+6、5p5p 和 8p+28p+2。

(a) 求常数 pp 的可能值。

(3)

(b) 在 (a) 中求出的 pp 值中,有一个是负分数。

利用这个 pp 值,求该等比数列的无穷项和。

(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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等比数列的中项平方等于两旁项的乘积。由此得到关于 pp 的二次方程。

答题过程

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The geometric condition gives

25p2=(2p+6)(8p+2)⟺9p2−52p−12=(9p+2)(p−6)=0.25p^2=(2p+6)(8p+2) \quad\Longleftrightarrow\quad 9p^2-52p-12=(9p+2)(p-6)=0.

Thus p=−2/9p=-2/9 or p=6p=6. The first and second terms are nonzero for both values, so the condition gives a geometric progression in either case.

(b)

解法一

思路

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用负的 pp 求首项和公比。公比的绝对值小于 11,所以可以求无穷和。

答题过程

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For p=−2/9p=-2/9, the first term is 50/950/9 and the second term is −10/9-10/9. The ratio is −1/5-1/5, so the sum to infinity is

50/91−(−1/5)=12527.\frac{50/9}{1-(-1/5)}=\frac{125}{27}.