AQA C2 2008 June — Question 3 7 marks

Exam BoardAQA
ModuleC2 (Core Mathematics 2)
Year2008
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeFind sum to infinity
DifficultyModerate -0.8 This is a straightforward geometric series question requiring only standard formula applications: finding r by division, applying S_∞ = a/(1-r), using S_n formula, and verifying the nth term formula. All parts are routine recall with no problem-solving or novel insight required, making it easier than average but not trivial due to the multi-part structure.
Spec1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1

3 A geometric series begins $$20 + 16 + 12.8 + 10.24 + \ldots$$
  1. Find the common ratio of the series.
  2. Find the sum to infinity of the series.
  3. Find the sum of the first 20 terms of the series, giving your answer to three decimal places.
  4. Prove that the \(n\)th term of the series is \(25 \times 0.8 ^ { n }\).

AnswerMarks Guidance
(a) \(r = 16 \div 20 = 0.8\)B1 (1 mark) OE
(b) \(\frac{a}{1-r} = \frac{20}{1-0.8} = 100\)M1, A1F (2 marks) OE Using a correct formula with \(a = 20\) or \(r = c's 0.8\); ft on \(c's\) value of \(r\) provided \(
(c) \(\{S_{20}\} = \frac{a(1-r^{20})}{1-r} = 100(1-0.8^{20}) = 98.847[07\ldots]\)M1, A1 (2 marks) OE Using a correct formula with \(n = 20\); Condone > 3dp
(d) \(nth\) term \(= 20 \cdot r^{n-1} = 20(0.8)^{n-1} = 20 \times 0.8^n \times 0.8^{-1} = 25 \times 0.8^n\)M1, A1 (2 marks) Ft on \(c's\) \(r\). Award even if \(16^{-1}\) seen; CSO; AG
Total for Q3: 7 marks
**(a)** $r = 16 \div 20 = 0.8$ | B1 (1 mark) | OE

**(b)** $\frac{a}{1-r} = \frac{20}{1-0.8} = 100$ | M1, A1F (2 marks) | OE Using a correct formula with $a = 20$ or $r = c's 0.8$; ft on $c's$ value of $r$ provided $|r| < 1$

**(c)** $\{S_{20}\} = \frac{a(1-r^{20})}{1-r} = 100(1-0.8^{20}) = 98.847[07\ldots]$ | M1, A1 (2 marks) | OE Using a correct formula with $n = 20$; Condone > 3dp

**(d)** $nth$ term $= 20 \cdot r^{n-1} = 20(0.8)^{n-1} = 20 \times 0.8^n \times 0.8^{-1} = 25 \times 0.8^n$ | M1, A1 (2 marks) | Ft on $c's$ $r$. Award even if $16^{-1}$ seen; CSO; AG

**Total for Q3: 7 marks**

---
3 A geometric series begins

$$20 + 16 + 12.8 + 10.24 + \ldots$$
\begin{enumerate}[label=(\alph*)]
\item Find the common ratio of the series.
\item Find the sum to infinity of the series.
\item Find the sum of the first 20 terms of the series, giving your answer to three decimal places.
\item Prove that the $n$th term of the series is $25 \times 0.8 ^ { n }$.
\end{enumerate}

\hfill \mbox{\textit{AQA C2 2008 Q3 [7]}}