Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2026 June Paper 11 Q1

A Level / CIE / P1

CIE 9709 2026 June Paper 11 Paper · Question 1

题目

Problem

In a geometric progression, the third term is 5050 and the fifth term is 12.512.5.

Given that the common ratio of the progression is positive, find the sum to infinity of the progression.

(4)
题目中文翻译

一个等比数列的第三项为 5050,第五项为 12.512.5。

已知该数列的公比为正,求其无穷和。

(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

解法一

思路

展开

第五项比第三项多乘两次公比。由两项之比确定正公比,再从第三项退回首项。

答题过程

展开

If aa is the first term and r>0r>0 is the common ratio, then ar2=50ar^2=50 and ar4=12.5ar^4=12.5. Dividing gives r2=12.5/50=1/4r^2=12.5/50=1/4, so r=1/2r=1/2 and a=50/(1/4)=200a=50/(1/4)=200. Since ∣r∣<1|r|<1, the sum to infinity is 200/(1−1/2)=400200/(1-1/2)=400.

解法二

思路

展开

第三项的平方除以第五项,公比的幂次正好抵消,可先直接求首项。

答题过程

展开

Write the third and fifth terms as ar2=50ar^2=50 and ar4=12.5ar^4=12.5. Dividing the square of the first equality by the second gives

a=(ar2)2ar4=50212.5=200.a=\frac{(ar^2)^2}{ar^4}=\frac{50^2}{12.5}=200.

Then r2=50/200=1/4r^2=50/200=1/4. Since r>0r>0, r=1/2r=1/2, which has absolute value less than 11. Hence the sum to infinity is

S∞=2001−1/2=400.S_\infty=\frac{200}{1-1/2}=400.