| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find N for S_∞ - S_N condition |
| Difficulty | Moderate -0.3 This is a straightforward geometric series question requiring standard formula application: finding a term using ar^(n-1), sum to infinity using a/(1-r), and solving S_n > 72 algebraically or by trial. The calculations are routine for C2 level with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u_n=ar^{n-1} \Rightarrow u_{25}=6\times0.92^{24}=\) awrt \(0.81\) | M1A1 | M1: attempts \(6\times0.92^{24}\) or \(6\times0.92^{25-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S_\infty=\frac{a}{1-r} \Rightarrow S_\infty=\frac{6}{1-0.92}=75\) | M1A1 | M1: attempts \(S_\infty=\frac{a}{1-r}\) with \(a=6\), \(r=0.92\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sets \(S_n>72 \Rightarrow \frac{6(1-0.92^n)}{1-0.92}>72\) | M1 | Accept \(=72\) or \(<72\) |
| \(0.92^n<0.04\) | A1 | Accept \(0.92^n=0.04\) |
| Takes logs: \(n>\frac{\log0.04}{\log0.92}\) | dM1 | Proceeds from \(a^n....b\) to \(n....\log_a b\) or \(n....\frac{\log b}{\log a}\); \(a\) and \(b\) must be positive |
| \(n=39\) | A1 | cso. Do not accept \(n=38.6\). Trial and improvement gains all 4 marks if \(n=39\) stated |
## Question 9:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u_n=ar^{n-1} \Rightarrow u_{25}=6\times0.92^{24}=$ awrt $0.81$ | M1A1 | M1: attempts $6\times0.92^{24}$ or $6\times0.92^{25-1}$ |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_\infty=\frac{a}{1-r} \Rightarrow S_\infty=\frac{6}{1-0.92}=75$ | M1A1 | M1: attempts $S_\infty=\frac{a}{1-r}$ with $a=6$, $r=0.92$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets $S_n>72 \Rightarrow \frac{6(1-0.92^n)}{1-0.92}>72$ | M1 | Accept $=72$ or $<72$ |
| $0.92^n<0.04$ | A1 | Accept $0.92^n=0.04$ |
| Takes logs: $n>\frac{\log0.04}{\log0.92}$ | dM1 | Proceeds from $a^n....b$ to $n....\log_a b$ or $n....\frac{\log b}{\log a}$; $a$ and $b$ must be positive |
| $n=39$ | A1 | cso. Do not accept $n=38.6$. Trial and improvement gains all 4 marks if $n=39$ stated |
---\begin{enumerate}
\item The first term of a geometric series is 6 and the common ratio is 0.92
\end{enumerate}
For this series, find\\
(a) (i) the $25 ^ { \text {th } }$ term, giving your answer to 2 significant figures,\\
(ii) the sum to infinity.
The sum to $n$ terms of this series is greater than 72\\
(b) Calculate the smallest possible value of $n$.\\
VJYV SIHI NITIIIUMION, OC\\
VILV SIHI NI JAHM ION OO\\
VILV SIHI NI JIIIM ION OO\\
\hfill \mbox{\textit{Edexcel C12 2016 Q9 [8]}}