Standard +0.8 This question requires understanding of infinite geometric series, deriving the sum to infinity formula, setting up an inequality involving the remainder term, and solving a logarithmic inequality to find N. While the geometric series itself is standard A-level content, the requirement to find N such that the tail sum is below a threshold involves multiple conceptual steps (recognizing the tail is itself a geometric series, manipulating the inequality, applying logarithms correctly with sign considerations) that go beyond routine exercises. This is moderately challenging but within reach of strong A-level students.
7 In this question you must show detailed reasoning.
A sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } \ldots\) is defined by \(u _ { n } = 25 \times 0.6 ^ { n }\).
Use an algebraic method to find the smallest value of \(N\) such that \(\sum _ { n = 1 } ^ { \infty } u _ { n } - \sum _ { n = 1 } ^ { N } u _ { n } < 10 ^ { - 4 }\).
Identify GP; correct \(a\) and \(r\) soi. Stated or implied by use in equation
\(S_\infty = \frac{15}{1-0.6}\)
B1
Correct \(S_\infty\) with their \(a\) and \(r\). Must be using correct formula. Allow \(a = 25\) even if not stated explicitly before formula is used. Allow \(a=15\), \(r=0.6\) and \(\frac{a}{1-r} = 37.5\) to imply B1. B0 for 37.5 with no evidence
\(S_N = \frac{15(1-0.6^N)}{1-0.6}\)
B1
Correct \(S_N\) with their \(a\) and \(r\). Must be using correct formula. Allow \(a = 25\) even if not stated explicitly
Link \(S_\infty - S_N\) to \(10^{-4}\) and attempt to rearrange. As far as \(p \times 0.6^N < q\). Condone either \(=\) or any inequality sign. M0 for e.g. \(15 \times 0.6^N = 9^N\) or \(1 - 0.6^N = 0.4^N\)
\(0.6^N < 2.67 \times 10^{-6}\)
A1
Correct equation in useable form. Any linking sign. If using logs on \(37.5 \times 0.6^N\) then product must be dealt with correctly
\(N > \log_{0.6}(2.67 \times 10^{-6})\)
M1
Use logs to solve equation. Either take logs on both sides (consistent base), drop power and rearrange, or take \(\log_{0.6}\) on RHS. Any linking sign including inequality sign that does not change direction
\(N > 25.125\ldots\)
A1
Obtain 25.1 / 25 / 26. Any sign. No evidence of use of logs — award B1 instead of M1A1
hence \(N = 26\)
A1
Obtain \(N = 26\) only (or e.g. \(N\) is 26) www. A0 if inequality e.g. \(N \geq 26\). A0 if from incorrect inequality e.g. \(N < 25.125\) unless recovered by testing
[8]
## Question 7:
| Answer | Mark | Guidance |
|--------|------|----------|
| **DR**. GP with $a = 15$, $r = 0.6$ | B1 | Identify GP; correct $a$ and $r$ soi. Stated or implied by use in equation |
| $S_\infty = \frac{15}{1-0.6}$ | B1 | Correct $S_\infty$ with their $a$ and $r$. Must be using correct formula. Allow $a = 25$ even if not stated explicitly before formula is used. Allow $a=15$, $r=0.6$ and $\frac{a}{1-r} = 37.5$ to imply B1. B0 for 37.5 with no evidence |
| $S_N = \frac{15(1-0.6^N)}{1-0.6}$ | B1 | Correct $S_N$ with their $a$ and $r$. Must be using correct formula. Allow $a = 25$ even if not stated explicitly |
| $37.5 - 37.5(1-0.6^N) < 10^{-4}$, $37.5 \times 0.6^N < 10^{-4}$ | M1 | Link $S_\infty - S_N$ to $10^{-4}$ and attempt to rearrange. As far as $p \times 0.6^N < q$. Condone either $=$ or any inequality sign. M0 for e.g. $15 \times 0.6^N = 9^N$ or $1 - 0.6^N = 0.4^N$ |
| $0.6^N < 2.67 \times 10^{-6}$ | A1 | Correct equation in useable form. Any linking sign. If using logs on $37.5 \times 0.6^N$ then product must be dealt with correctly |
| $N > \log_{0.6}(2.67 \times 10^{-6})$ | M1 | Use logs to solve equation. Either take logs on both sides (consistent base), drop power and rearrange, or take $\log_{0.6}$ on RHS. Any linking sign including inequality sign that does not change direction |
| $N > 25.125\ldots$ | A1 | Obtain 25.1 / 25 / 26. Any sign. No evidence of use of logs — award B1 instead of M1A1 |
| hence $N = 26$ | A1 | Obtain $N = 26$ only (or e.g. $N$ is 26) www. A0 if inequality e.g. $N \geq 26$. A0 if from incorrect inequality e.g. $N < 25.125$ unless recovered by testing |
| **[8]** | | |
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7 In this question you must show detailed reasoning.
A sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } \ldots$ is defined by $u _ { n } = 25 \times 0.6 ^ { n }$.\\
Use an algebraic method to find the smallest value of $N$ such that $\sum _ { n = 1 } ^ { \infty } u _ { n } - \sum _ { n = 1 } ^ { N } u _ { n } < 10 ^ { - 4 }$.
\hfill \mbox{\textit{OCR H240/01 2019 Q7 [8]}}