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IAL 2022 Oct Q8

A Level / Edexcel / P2

IAL 2022 Oct Paper · Question 8

题目

Problem

A geometric sequence has first term aa and common ratio rr

Given that S∞=3aS_{\infty} = 3a

(a) show that r=23r = \dfrac{2}{3}

(2)

Given also that

u2−u4=16u_2 - u_4 = 16

where uku_k is the kthk\text{th} term of this sequence,

(b) find the value of S10S_{10} giving your answer to one decimal place.

(5)

解答

(a)

解法一

思路

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等比数列（Geometric sequence）的无穷项和（Sum to infinity）公式为：

S∞=a1−r(∣r∣<1)S_{\infty} = \frac{a}{1 - r} \qquad (|r| < 1)

已知 S∞=3aS_{\infty} = 3a，我们可以建立方程：

a1−r=3a\frac{a}{1 - r} = 3a

因为等比数列的首项 a≠0a \neq 0（否则数列恒为 0，没有意义且无穷和也为 0），我们可以在方程两边同时约去 aa：

11−r=3\frac{1}{1 - r} = 3

解这个关于 rr 的一元一次方程即可：

1−r=13  ⟹  r=1−13=231 - r = \frac{1}{3} \implies r = 1 - \frac{1}{3} = \frac{2}{3}

答题过程

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Using the formula for the sum to infinity of a geometric sequence:

S∞=a1−rS_{\infty} = \frac{a}{1 - r}

Given that S∞=3aS_{\infty} = 3a:

a1−r=3a\frac{a}{1 - r} = 3a

Since a≠0a \neq 0, we can divide both sides by aa:

11−r=  31−r=  13r=  1−13r=  23(proven)\begin{align*} \frac{1}{1 - r} =&\,\, 3 \\[2mm] 1 - r =&\,\, \frac{1}{3} \\[2mm] r =&\,\, 1 - \frac{1}{3} \\[2mm] r =&\,\, \frac{2}{3} \quad \text{(proven)} \end{align*}

(b)

解法一

思路

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等比数列的通项公式为 uk=ark−1u_k = a r^{k-1}。 因此我们可以将 u2u_2 和 u4u_4 用首项 aa 和公比 rr 表达：

  • u2=aru_2 = ar
  • u4=ar3u_4 = ar^3 已知 u2−u4=16u_2 - u_4 = 16，代入可得：
ar−ar3=16  ⟹  ar(1−r2)=16ar - ar^3 = 16 \implies ar(1 - r^2) = 16

将 (a) 中求得的 r=23r = \dfrac{2}{3} 代入方程：

a(23)(1−(23)2)=16  ⟹  a(23)(1−49)=16  ⟹  a(23)(59)=16a \left(\frac{2}{3}\right) \left(1 - \left(\frac{2}{3}\right)^2\right) = 16 \implies a \left(\frac{2}{3}\right) \left(1 - \frac{4}{9}\right) = 16 \implies a \left(\frac{2}{3}\right) \left(\frac{5}{9}\right) = 16 a⋅1027=16  ⟹  a=16×2710=43.2a \cdot \frac{10}{27} = 16 \implies a = 16 \times \frac{27}{10} = 43.2

现在我们拥有首项 a=43.2a = 43.2，公比 r=23r = \dfrac{2}{3}。需要计算前 10 项和 S10S_{10}，使用前 nn 项和公式：

Sn=a(1−rn)1−rS_n = \frac{a(1 - r^n)}{1 - r}

代入 n=10n = 10：

S10=43.2(1−(2/3)10)1−2/3=3×43.2(1−(23)10)=129.6(1−102459049)S_{10} = \frac{43.2 \left(1 - (2/3)^{10}\right)}{1 - 2/3} = 3 \times 43.2 \left(1 - \left(\frac{2}{3}\right)^{10}\right) = 129.6 \left(1 - \frac{1024}{59049}\right)

利用计算器计算并保留一位小数。

答题过程

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The general term of a geometric sequence is given by uk=ark−1u_k = ar^{k-1}.

Therefore:

u2=ar,u4=ar3u_2 = ar, \quad u_4 = ar^3

Given that u2−u4=16u_2 - u_4 = 16:

ar−ar3=  16ar(1−r2)=  16\begin{align*} ar - ar^3 =&\,\, 16 \\[2mm] ar(1 - r^2) =&\,\, 16 \end{align*}

Substitute r=23r = \dfrac{2}{3} into the equation:

a(23)(1−(23)2)=  16a(23)(1−49)=  16a(23)(59)=  161027a=  16a=  16×2710a=  43.2(or 2165)\begin{align*} a\left(\frac{2}{3}\right)\left(1 - \left(\frac{2}{3}\right)^2\right) =&\,\, 16 \\[2mm] a\left(\frac{2}{3}\right)\left(1 - \frac{4}{9}\right) =&\,\, 16 \\[2mm] a\left(\frac{2}{3}\right)\left(\frac{5}{9}\right) =&\,\, 16 \\[2mm] \frac{10}{27}a =&\,\, 16 \\[2mm] a =&\,\, 16 \times \frac{27}{10} \\[2mm] a =&\,\, 43.2 \quad \left(\text{or } \frac{216}{5}\right) \end{align*}

Now, using the sum formula Sn=a(1−rn)1−rS_n = \dfrac{a(1 - r^n)}{1 - r} to find S10S_{10}:

S10=  43.2(1−(23)10)1−23=  43.2(1−(23)10)13=  129.6(1−(23)10)=  129.6(1−102459049)=  129.6×0.982658...=  127.352...≈  127.4(to 1 decimal place)\begin{align*} S_{10} =&\,\, \frac{43.2\left(1 - \left(\frac{2}{3}\right)^{10}\right)}{1 - \frac{2}{3}} \\[2mm] =&\,\, \frac{43.2\left(1 - \left(\frac{2}{3}\right)^{10}\right)}{\frac{1}{3}} \\[2mm] =&\,\, 129.6\left(1 - \left(\frac{2}{3}\right)^{10}\right) \\[2mm] =&\,\, 129.6\left(1 - \frac{1024}{59049}\right) \\[2mm] =&\,\, 129.6 \times 0.982658... \\[2mm] =&\,\, 127.352... \\[2mm] \approx&\,\, 127.4 \quad (\text{to 1 decimal place}) \end{align*}