| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find year when threshold exceeded |
| Difficulty | Moderate -0.3 This is a straightforward application of standard geometric sequence formulas (nth term and sum) with calculator work. Part (b) requires trial-and-error or logarithms to find n, which adds minimal challenge. Slightly easier than average due to being purely procedural with no conceptual obstacles. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(t_{20} = 5 \times 1.2^{19} = 159.7\) | M1A1 | M1: Use of \(t_n = ar^{n-1}\). A1: Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S_{20} = \frac{5(1 - 1.2^{20})}{1 - 1.2} = 933.4\) | M1A1 | M1: Use of correct sum formula with \(n=19\) or \(n=20\). NB if \(n=19\) used and no formula quoted, score M0. A1: Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{5(1-1.2^n)}{1-1.2} (> \text{ or } =) 3000\) | B1 | Correct statement (allow '\(a\)' and/or '\(r\)' instead of 5 and 1.2) |
| \(1.2^n > 121\) | M1 | \(1.2^n\, (>\text{ or } < \text{ or } =)\, k\) |
| \(\log 1.2^n > \log 121\) or \(n > \log_{1.2} 121\) | M1 | Takes logs correctly |
| \(n > \frac{\log 121}{\log 1.2}\) i.e. \(n = 27\) | A1 | Cao |
## Question 4:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $t_{20} = 5 \times 1.2^{19} = 159.7$ | M1A1 | M1: Use of $t_n = ar^{n-1}$. A1: Cao |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_{20} = \frac{5(1 - 1.2^{20})}{1 - 1.2} = 933.4$ | M1A1 | M1: Use of correct sum formula with $n=19$ or $n=20$. NB if $n=19$ used and no formula quoted, score M0. A1: Cao |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{5(1-1.2^n)}{1-1.2} (> \text{ or } =) 3000$ | B1 | Correct statement (allow '$a$' and/or '$r$' instead of 5 and 1.2) |
| $1.2^n > 121$ | M1 | $1.2^n\, (>\text{ or } < \text{ or } =)\, k$ |
| $\log 1.2^n > \log 121$ or $n > \log_{1.2} 121$ | M1 | Takes logs correctly |
| $n > \frac{\log 121}{\log 1.2}$ i.e. $n = 27$ | A1 | Cao |
---4. The first term of a geometric series is 5 and the common ratio is 1.2
For this series find, to 1 decimal place,
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item the $20 ^ { \text {th } }$ term,
\item the sum of the first 20 terms.
The sum of the first $n$ terms of the series is greater than 3000
\end{enumerate}\item Calculate the smallest possible value of $n$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q4 [8]}}