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CIE 9709 2022 November Paper 12 Q2

A Level / CIE / P1

CIE 9709 2022 Nov Paper 12 Paper · Question 2

题目

Problem

The first, second and third terms of an arithmetic progression are aa, 2a2a and a2a^2 respectively, where aa is a positive constant.

Find the sum of the first 50 terms of the progression.

(5)
题目中文翻译

一个等差数列的第一、第二和第三项分别为 aa、2a2a 和 a2a^2,其中 aa 为正常数。

求该数列前 50 项的和。

(5)
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

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等差数列的两个相邻差相等。用 a>0a>0 选出首项,再算前 5050 项和。

答题过程

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The common difference satisfies 2a−a=a2−2a2a-a=a^2-2a, so a(a−3)=0a(a-3)=0. Since a>0a>0, we have a=3a=3 and d=3d=3. Hence

S50=502(2(3)+49(3))=3825.S_{50}=\frac{50}{2}\bigl(2(3)+49(3)\bigr)=3825.