| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find N for S_∞ - S_N condition |
| Difficulty | Standard +0.8 This is a multi-part geometric series question that progresses from routine calculations (parts a,b) to algebraic manipulation requiring logarithms (part c) and interpretation (part d). The proof in part (c) requires students to manipulate the sum formula, set up an inequality, and apply logarithm laws correctly—going beyond standard textbook exercises. However, it's still within C2 scope with clear scaffolding, making it moderately challenging but not exceptional. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(T_{20} = 5 \times \left(\frac{4}{5}\right)^{19} = 0.072\) (Accept awrt) | M1 A1 (2) |
| (b) | \(S_\infty = \frac{5}{1-0.8} = 25\) | M1 A1 (2) |
| (c) | \(\frac{5(1-0.8^k)}{1-0.8} > 24.95\) | M1 |
| \(1 - 0.8^k > 0.998\) (or equivalent) | A1 | Allow with \(=\) or \(<\) |
| \(k\log 0.8 < \log 0.002\) or \(k > \log_{0.8} 0.002\) | M1 | Allow with \(=\) or \(<\); M1: introducing logs and using laws of logs correctly, dealing with power \(k\) so that \(p^k = k\log p\) |
| \(k > \frac{\log 0.002}{\log 0.8}\) (*) | A1cso (4) | 2nd A1: incorrect statement (including equalities) at any stage loses this mark |
| (d) | \(k = 28\) (must be integer value) | B1 (1) |
## Question 6:
**(a)** | $T_{20} = 5 \times \left(\frac{4}{5}\right)^{19} = 0.072$ (Accept awrt) | M1 A1 (2) | M1: requires use of correct formula $ar^{n-1}$; allow $5 \times \frac{4^{19}}{5}$ for M1 |
**(b)** | $S_\infty = \frac{5}{1-0.8} = 25$ | M1 A1 (2) | M1: requires use of correct formula $\frac{a}{1-r}$ |
**(c)** | $\frac{5(1-0.8^k)}{1-0.8} > 24.95$ | M1 | Allow with $=$ or $<$ |
| $1 - 0.8^k > 0.998$ (or equivalent) | A1 | Allow with $=$ or $<$ |
| $k\log 0.8 < \log 0.002$ or $k > \log_{0.8} 0.002$ | M1 | Allow with $=$ or $<$; M1: introducing logs and using laws of logs correctly, dealing with power $k$ so that $p^k = k\log p$ |
| $k > \frac{\log 0.002}{\log 0.8}$ (*) | A1cso (4) | 2nd A1: incorrect statement (including equalities) at any stage loses this mark |
**(d)** | $k = 28$ (must be integer value) | B1 (1) | Not $k > 27$, or $k < 28$, or $k > 28$ |
---6. A geometric series has first term 5 and common ratio $\frac { 4 } { 5 }$.
Calculate
\begin{enumerate}[label=(\alph*)]
\item the 20th term of the series, to 3 decimal places,
\item the sum to infinity of the series.
Given that the sum to $k$ terms of the series is greater than 24.95,
\item show that $k > \frac { \log 0.002 } { \log 0.8 }$,
\item find the smallest possible value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q6 [9]}}