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CIE 9709 2022 June Paper 11 Q2

A Level / CIE / P1

CIE 9709 2022 June Paper 11 Paper · Question 2

题目

Problem

The thirteenth term of an arithmetic progression is 1212 and the sum of the first 3030 terms is −15-15.

Find the sum of the first 5050 terms of the progression.

(5)
题目中文翻译

一个等差数列的第十三项为 1212,前 3030 项的和为 −15-15。

求这个数列的前 5050 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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把第 1313 项和前 3030 项和分别写成首项、公差的方程。解出两者后再求前 5050 项和。

答题过程

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Let the first term be aa and the common difference be dd. The conditions give

 a+12d=12, 302(2a+29d)=−15.\begin{align*} &\,a+12d=12,\qquad\\ &\,\frac{30}{2}(2a+29d)=-15. \end{align*}

Thus 2a+29d=−12a+29d=-1. Subtracting twice the first equation gives 5d=−255d=-25, so d=−5d=-5 and a=72a=72. Therefore

S50=502(2(72)+49(−5))=−2525.S_{50}=\frac{50}{2}(2(72)+49(-5))=-2525.