Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2022 June Paper 13 Q3

A Level / CIE / P1

CIE 9709 2022 June Paper 13 Paper · Question 3

题目

Problem

An arithmetic progression has first term 44 and common difference dd. The sum of the first nn terms of the progression is 58635863.

(a) Show that (n−1)d=11726n−8(n-1)d=\frac{11726}{n}-8.

(1)

(b) Given that the nnth term is 139139, find the values of nn and dd, giving the value of dd as a fraction.

(4)
题目中文翻译

一个等差数列的首项为 44,公差为 dd。这个数列的前 nn 项的和为 58635863。

(a) 证明 (n−1)d=11726n−8(n-1)d=\frac{11726}{n}-8。

(1)

(b) 已知第 nn 项为 139139,求 nn 和 dd 的值,并将 dd 的值写成分数。

(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

(a)

解法一

思路

展开

将首项和已知总和代入等差求和公式,再孤立含公差的一项。

答题过程

展开

We have

5863=n2(8+(n−1)d).5863=\frac n2\bigl(8+(n-1)d\bigr).

Multiplying by 22 gives

n[8+(n−1)d]=11726.n\Big[8+(n-1)d\Big]=11726.

Since n>0n>0, division by nn and subtraction of 88 give (n−1)d=11726/n−8(n-1)d=11726/n-8.

(b)

解法一

思路

展开

先由末项求出 (n−1)d(n-1)d,再用 (a) 确定 nn。

答题过程

展开

The last term is 4+(n−1)d=1394+(n-1)d=139, so (n−1)d=135(n-1)d=135. Substitution in (a) gives 135=11726/n−8135=11726/n-8, hence n=11726/143=82n=11726/143=82. Therefore d=135/81=5/3d=135/81=5/3.