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CIE 9709 2024 June Paper 13 Q10

A Level / CIE / P1

CIE 9709 2024 June Paper 13 Paper · Question 10

题目

Problem

The geometric progression a1,a2,a3,…a_1, a_2, a_3, \ldots has first term 22 and common ratio rr where r>0r > 0. It is given that 92a5+7a3=8\frac{9}{2}a_5 + 7a_3 = 8.

(a) Find the value of rr.

[3]

(b) Find the sum of the first 2020 terms of the geometric progression. Give your answer correct to 44 significant figures.

[2]

(c) Find the sum to infinity of the progression a2,a5,a8,…a_2, a_5, a_8, \ldots.

[3]
题目中文翻译

等比数列 a1,a2,a3,…a_1, a_2, a_3, \ldots 的首项为 22,公比为 rr,其中 r>0r > 0。已知 92a5+7a3=8\frac{9}{2}a_5 + 7a_3 = 8。

(a) 求 rr 的值。

[3]

(b) 求该等比数列前 2020 项的和。答案保留 44 位有效数字。

[2]

(c) 求数列 a2,a5,a8,…a_2, a_5, a_8, \ldots 的无穷和。

[3]

解答

(a)

解法一

思路

展开

用 an=2rn−1a_n=2r^{n-1} 写第三项和第五项。得到关于 r2r^2 的二次方程后,只保留正的公比。

答题过程

展开

The condition gives 9r4+14r2−8=09r^4+14r^2-8=0. Putting u=r2u=r^2 gives

9u2+14u−8=(9u−4)(u+2)=0.9u^2+14u-8=(9u-4)(u+2)=0.

As u>0u>0, u=4/9u=4/9. The positive ratio is r=2/3r=2/3.

(b)

解法一

思路

展开

用等比数列有限和公式,并保留精确公比到最后。

答题过程

展开 S20= 2(1−(2/3)20)1−2/3= 6(1−(2/3)20)≈5.998\begin{align*} S_{20}=&\,\frac{2(1-(2/3)^{20})}{1-2/3}\\ =&\,6\bigl(1-(2/3)^{20}\bigr)\approx5.998 \end{align*}

to four significant figures.

(c)

解法一

思路

展开

新数列从原第二项开始,每次跨过三项的位置差,所以新公比是 r3r^3。

答题过程

展开

The first term is a2=4/3a_2=4/3, and the common ratio is (2/3)3=8/27(2/3)^3=8/27. It lies between 00 and 11, so the sum to infinity is

4/31−8/27=3619.\frac{4/3}{1-8/27}=\frac{36}{19}.