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CIE 9709 2024 March Paper 12 Q8

A Level / CIE / P1

CIE 9709 2024 March Paper 12 Paper · Question 8

题目

Problem

(a) An arithmetic progression is such that its first term is 6 and its tenth term is 19.5.

Find the sum of the first 100 terms of this arithmetic progression.

[4]

(b) A geometric progression a1,a2,a3,…a_1, a_2, a_3, \ldots is such that a1=24a_1 = 24 and the common ratio is 12\frac{1}{2}.

The sum to infinity of this geometric progression is denoted by SS. The sum to infinity of the even-numbered terms (i.e. a2,a4,a6,…a_2, a_4, a_6, \ldots) is denoted by SES_E.

Find the values of SS and SES_E.

[4]
题目中文翻译

(a) 一个等差数列的首项为 6,第 10 项为 19.5。

求该等差数列前 100 项的和。

[4]

(b) 一个等比数列 a1,a2,a3,…a_1, a_2, a_3, \ldots 满足 a1=24a_1 = 24,公比为 12\frac{1}{2}。

该等比数列的无穷和记为 SS。偶数项(即 a2,a4,a6,…a_2, a_4, a_6, \ldots)的无穷和记为 SES_E。

求 SS 和 SES_E 的值。

[4]

解答

(a)

解法一

思路

展开

首项到第十项相差九个公差。求出公差后,用等差求和公式求前一百项的和。

答题过程

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We have 6+9d=19.56+9d=19.5, so d=1.5d=1.5. Then

S100=1002(2(6)+99(1.5))=8025.S_{100}=\frac{100}{2}\bigl(2(6)+99(1.5)\bigr)=8025.

(b)

解法一

思路

展开

偶数项本身形成新等比数列。首项是原来的第二项,之后每次跨两项,所以公比要平方。

答题过程

展开

For the full progression,

S=241−1/2=48.S=\frac{24}{1-1/2}=48.

The even-numbered terms start at 24(1/2)=1224(1/2)=12 and have ratio (1/2)2=1/4(1/2)^2=1/4. Hence

SE=121−1/4=16.S_E=\frac{12}{1-1/4}=16.