Question 9 - A-Level Maths

Edexcel C2 2006 June — Question 9 11 marks

Exam Board	Edexcel
Module	C2 (Core Mathematics 2)
Year	2006
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Geometric Sequences and Series
Type	Relationship between two GPs
Difficulty	Moderate -0.3 This is a structured multi-part question testing standard geometric series formulas (sum to infinity, nth term) with straightforward algebraic manipulation. While it requires multiple steps and connecting several concepts, each part follows directly from textbook methods with no novel insight needed. The quadratic solving and logarithm work in part (e) are routine C2 techniques, making this slightly easier than average.
Spec	1.04i Geometric sequences: nth term and finite series sum 1.04j Sum to infinity: convergent geometric series \|r\|<1

A geometric series has first term $a$ and common ratio $r$. The second term of the series is 4 and the sum to infinity of the series is 25.
1. Show that $25 r ^ { 2 } - 25 r + 4 = 0$.
2. Find the two possible values of $r$.
3. Find the corresponding two possible values of $a$.
4. Show that the sum, $S _ { n }$, of the first $n$ terms of the series is given by
$$S _ { n } = 25 \left( 1 - r ^ { n } \right) .$$ Given that $r$ takes the larger of its two possible values,
find the smallest value of $n$ for which $S _ { n }$ exceeds 24 .

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) $ar = 4$, $\frac{a}{1 - r} = 25$ (These can be seen elsewhere)	B1, B1
$a = 25(1 - r)$	M1	$25r(1 - r) = 4$
$25r^2 - 25r + 4 = 0$	A1cso	(4 marks)
(b) $(5r - 1)(5r - 4) = 0$	M1, A1	$r = \ldots$, $\frac{1}{5}$ or $\frac{4}{5}$ (2 marks)
(c) $r = \ldots \Rightarrow a = \ldots$,	M1, A1	$20$ or $5$ (2 marks)
(d) $S_n = \frac{a(1 - r^n)}{1 - r}$, but $\frac{a}{1 - r} = 25$, so	B1	$S_n = 25(1 - r^n)$ (*) (1 mark)
(e) $25(1 - 0.8^n) > 24$ and proceed to $n = \ldots$ (or $>, $ or $<$) with no unsound algebra.	M1	$n > \log 0.04 / \log 0.8$ (= 14.425...)
(11 marks)	(a) The M mark is not dependent, but both expressions must contain both $a$ and $r$. (b) Special case: One correct $r$ value given, with no method (or perhaps trial and error): B1 B0. (c) M1: Substitute one $r$ value back to find a value of $a$. (d) Sufficient here to verify with just one pair of values of $a$ and $r$. (e) Accept "=" rather than inequalities throughout, and also allow the wrong inequality to be used at any stage. M1 requires use of their larger value of $r$. A correct answer with no working scores both marks. For "trial and error" methods, to score M1, a value of $n$ between 12 and 18 (inclusive) must be tried.

**(a)** $ar = 4$, $\frac{a}{1 - r} = 25$ (These can be seen elsewhere) | B1, B1 | |
| --- | --- | --- |
| $a = 25(1 - r)$ | M1 | $25r(1 - r) = 4$ |
| $25r^2 - 25r + 4 = 0$ | A1cso | (4 marks) |
| **(b)** $(5r - 1)(5r - 4) = 0$ | M1, A1 | $r = \ldots$, $\frac{1}{5}$ or $\frac{4}{5}$ (2 marks) |
| **(c)** $r = \ldots \Rightarrow a = \ldots$, | M1, A1 | $20$ or $5$ (2 marks) |
| **(d)** $S_n = \frac{a(1 - r^n)}{1 - r}$, but $\frac{a}{1 - r} = 25$, so | B1 | $S_n = 25(1 - r^n)$ (*) (1 mark) |
| **(e)** $25(1 - 0.8^n) > 24$ and proceed to $n = \ldots$ (or $>, $ or $<$) with no unsound algebra. | M1 | $n > \log 0.04 / \log 0.8$ (= 14.425...) | A1 | $n = 15$ (2 marks) |
| | (11 marks) | **(a)** The M mark is not dependent, but both expressions must contain both $a$ and $r$. **(b)** Special case: One correct $r$ value given, with no method (or perhaps trial and error): B1 B0. **(c)** M1: Substitute one $r$ value back to find a value of $a$. **(d)** Sufficient here to verify with just one pair of values of $a$ and $r$. **(e)** Accept "=" rather than inequalities throughout, and also allow the wrong inequality to be used at any stage. M1 requires use of their larger value of $r$. A correct answer with no working scores both marks. For "trial and error" methods, to score M1, a value of $n$ between 12 and 18 (inclusive) must be tried. |

Show LaTeX source

\begin{enumerate}
  \item A geometric series has first term $a$ and common ratio $r$. The second term of the series is 4 and the sum to infinity of the series is 25.\\
(a) Show that $25 r ^ { 2 } - 25 r + 4 = 0$.\\
(b) Find the two possible values of $r$.\\
(c) Find the corresponding two possible values of $a$.\\
(d) Show that the sum, $S _ { n }$, of the first $n$ terms of the series is given by
\end{enumerate}

$$S _ { n } = 25 \left( 1 - r ^ { n } \right) .$$

Given that $r$ takes the larger of its two possible values,\\
(e) find the smallest value of $n$ for which $S _ { n }$ exceeds 24 .\\

\hfill \mbox{\textit{Edexcel C2 2006 Q9 [11]}}

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Q1 4 Q2 5 Q3 4 Q4 8 Q5 8 Q6 4 Q7 8 Q8 9 Q9 11 Q10 14

Answer	Marks	Guidance
(a) \(ar = 4\), \(\frac{a}{1 - r} = 25\) (These can be seen elsewhere)	B1, B1
\(a = 25(1 - r)\)	M1	\(25r(1 - r) = 4\)
\(25r^2 - 25r + 4 = 0\)	A1cso	(4 marks)
(b) \((5r - 1)(5r - 4) = 0\)	M1, A1	\(r = \ldots\), \(\frac{1}{5}\) or \(\frac{4}{5}\) (2 marks)
(c) \(r = \ldots \Rightarrow a = \ldots\),	M1, A1	\(20\) or \(5\) (2 marks)
(d) \(S_n = \frac{a(1 - r^n)}{1 - r}\), but \(\frac{a}{1 - r} = 25\), so	B1	\(S_n = 25(1 - r^n)\) (*) (1 mark)
(e) \(25(1 - 0.8^n) > 24\) and proceed to \(n = \ldots\) (or \(>, \) or \(<\)) with no unsound algebra.	M1	\(n > \log 0.04 / \log 0.8\) (= 14.425...)
(11 marks)	(a) The M mark is not dependent, but both expressions must contain both \(a\) and \(r\). (b) Special case: One correct \(r\) value given, with no method (or perhaps trial and error): B1 B0. (c) M1: Substitute one \(r\) value back to find a value of \(a\). (d) Sufficient here to verify with just one pair of values of \(a\) and \(r\). (e) Accept "=" rather than inequalities throughout, and also allow the wrong inequality to be used at any stage. M1 requires use of their larger value of \(r\). A correct answer with no working scores both marks. For "trial and error" methods, to score M1, a value of \(n\) between 12 and 18 (inclusive) must be tried.