Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2026 June Paper 15 Q12

A Level / CIE / P1

CIE 9709 2026 June Paper 15 Paper · Question 12

题目

Problem

(a) The first term of an arithmetic progression is 57305730 and the common difference is −23-23.

Find the value of the first negative term.

(3)

(b) The first term of a geometric progression is sin⁡θ\sin\theta and the second term is 12tan⁡θ\frac{1}{2}\tan\theta, where 0<θ<12π0<\theta<\frac{1}{2}\pi.

(i) Find the set of values of θ\theta for which the progression is convergent.

(3)

(ii) Given that θ=16π\theta=\frac{1}{6}\pi, find the sum to infinity of the progression. Give your answer in the form p+qr\frac{p+\sqrt{q}}{r}, where pp, qq and rr are integers to be found.

(3)
题目中文翻译

(a) 一个等差数列的首项为 57305730,公差为 −23-23。

求第一个负数项的值。

(3)

(b) 一个等比数列的首项为 sin⁡θ\sin\theta,第二项为 12tan⁡θ\frac{1}{2}\tan\theta,其中 0<θ<12π0<\theta<\frac{1}{2}\pi。

(i) 求使该数列收敛的 θ\theta 的取值集合。

(3)

(ii) 已知 θ=16π\theta=\frac{1}{6}\pi,求该数列的无穷和。答案写成 p+qr\frac{p+\sqrt{q}}{r} 的形式,其中 pp、qq 和 rr 是待求的整数。

(3)
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

展开

写出通项,并用严格不等式确定最早出现负数的项号。

答题过程

展开

The nnth term is 5730−23(n−1)5730-23(n-1). It is negative exactly when n−1>5730/23=249+3/23n-1>5730/23=249+3/23. The smallest possible n−1n-1 is 250250, so the first negative term is 5730−23(250)=−205730-23(250)=-20.

(b)(i)

解法一

思路

展开

用第二项除以第一项求公比。在给定角度范围内公比为正,只需解公比小于 11 的条件。

答题过程

展开

The common ratio is

r=12tan⁡θsin⁡θ=12cos⁡θ.r=\frac{\tfrac12\tan\theta}{\sin\theta}=\frac1{2\cos\theta}.

For 0<θ<π/20<\theta<\pi/2, this is positive, so convergence requires r<1r<1. Equivalently cos⁡θ>1/2\cos\theta>1/2, which gives 0<θ<π/30<\theta<\pi/3.

(b)(ii)

解法一

思路

展开

代入指定角度求首项和公比,再用无穷等比和公式化成题设形式。

答题过程

展开

For θ=π/6\theta=\pi/6, the first term is 1/21/2 and the ratio is 1/31/\sqrt3. Thus

S∞= 1/21−1/3= 32(3−1)×3+13+1= 3+34.\begin{align*} S_\infty=&\,\frac{1/2}{1-1/\sqrt3}\\ =&\,\frac{\sqrt3}{2(\sqrt3-1)}\times\frac{\sqrt3+1}{\sqrt3+1}\\ =&\,\frac{3+\sqrt3}{4}. \end{align*}

The integers in the requested form are p=3p=3, q=3q=3 and r=4r=4.