Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2022 November Paper 13 Q9

A Level / CIE / P1

CIE 9709 2022 Nov Paper 13 Paper · Question 9

题目

Problem

The first term of a geometric progression is 216 and the fourth term is 64.

(a) Find the sum to infinity of the progression.

(3)

The second term of the geometric progression is equal to the second term of an arithmetic progression.

The third term of the geometric progression is equal to the fifth term of the same arithmetic progression.

(b) Find the sum of the first 21 terms of the arithmetic progression.

(6)
题目中文翻译

一个等比数列的第一项为 216,第四项为 64。

(a) 求该数列的无穷项和。

(3)

该等比数列的第二项等于一个等差数列的第二项。

该等比数列的第三项等于同一个等差数列的第五项。

(b) 求这个等差数列前 21 项的和。

(6)
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

从第一项到第四项经过三次乘公比。求出公比后检查是否收敛。

答题过程

展开

The common ratio satisfies 216r3=64216r^3=64, so r=2/3r=2/3. Since ∣r∣<1|r|<1, the sum to infinity is 216/(1−2/3)=648216/(1-2/3)=648.

(b)

解法一

思路

展开

等比数列的第二、三项给出等差数列的第二、五项。它们相隔三个公差。

答题过程

展开

The geometric terms are 216(2/3)=144216(2/3)=144 and 216(2/3)2=96216(2/3)^2=96. For the arithmetic progression,

a+d=144,a+4d=96.a+d=144,\qquad a+4d=96.

Thus d=−16d=-16 and a=160a=160. Its twenty-first term is 160+20(−16)=−160160+20(-16)=-160, so the sum of its first 2121 terms is 21(160−160)/2=021(160-160)/2=0.