| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2018 |
| Session | Specimen |
| 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 application of standard geometric series formulas with no conceptual challenges. Part (a) uses S_∞ = a/(1-r), part (b) applies S_N = a(1-r^N)/(1-r), and part (c) requires trial-and-error or logarithms to solve an inequality—all routine procedures for P2 level, 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 |
|---|---|---|
| \(S_{\infty} = \frac{20}{1-\frac{7}{8}} = 160\) | M1 | Use of a correct \(S_{\infty}\) formula |
| \(160\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{12} = \frac{20(1-(\frac{7}{8})^{12})}{1-\frac{7}{8}} = 127.77324... = 127.8\) (1 dp) | M1 A1 | M1: Use of correct \(S_n\) formula with \(n=12\) (condone missing brackets around \(\frac{7}{8}\)); A1: awrt 127.8 |
| Answer | Marks | Guidance |
|---|---|---|
| \(160 - \frac{20(1-(\frac{7}{8})^N)}{1-\frac{7}{8}} < 0.5\) | M1 | Applies \(S_N\) (GP only) with \(a=20\), \(r=\frac{7}{8}\) and uses 0.5 and their \(S_{\infty}\) at any point |
| \(160\left(\frac{7}{8}\right)^N < (0.5)\) or \(\left(\frac{7}{8}\right)^N < \left(\frac{0.5}{160}\right)\) | dM1 | Attempt to isolate \(+160\left(\frac{7}{8}\right)^N\) or \(\left(\frac{7}{8}\right)^N\) |
| \(N\log\left(\frac{7}{8}\right) < \log\left(\frac{0.5}{160}\right)\) | M1 | Uses law of logarithms to obtain equation or inequality of form \(N\log\left(\frac{7}{8}\right) < \log\left(\frac{0.5}{\text{their } S_{\infty}}\right)\) or \(N > \log_{0.875}\left(\frac{0.5}{\text{their } S_{\infty}}\right)\) |
| \(N > \frac{\log(\frac{0.5}{160})}{\log(\frac{7}{8})} = 43.19823...\) cso \(\Rightarrow N = 44\) | A1 cso | \(N=44\) (Allow \(N \geq 44\) but no \(N > 44\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Attempts \(160 - S_N\) or \(S_N\) with at least one value for \(N > 40\) | M1 | |
| Attempts \(160 - S_N\) or \(S_N\) with \(N=43\) or \(N=44\) | dM1 | |
| Evidence of examining \(160 - S_N\) or \(S_N\) for both \(N=43\) and \(N=44\) with both values correct to 2 DP; e.g. \(160 - S_{43}\) = awrt 0.51 and \(160 - S_{44}\) = awrt 0.45 or \(S_{43}\) = awrt 159.49 and \(S_{44}\) = awrt 159.55 | M1 | |
| \(N = 44\) | A1 cso | Answer of \(N=44\) only with no working scores no marks |
## Question 2:
### Part (a):
| $S_{\infty} = \frac{20}{1-\frac{7}{8}} = 160$ | M1 | Use of a correct $S_{\infty}$ formula |
|---|---|---|
| $160$ | A1 | |
### Part (b):
| $S_{12} = \frac{20(1-(\frac{7}{8})^{12})}{1-\frac{7}{8}} = 127.77324... = 127.8$ (1 dp) | M1 A1 | M1: Use of correct $S_n$ formula with $n=12$ (condone missing brackets around $\frac{7}{8}$); A1: awrt 127.8 |
|---|---|---|
### Part (c):
| $160 - \frac{20(1-(\frac{7}{8})^N)}{1-\frac{7}{8}} < 0.5$ | M1 | Applies $S_N$ (GP only) with $a=20$, $r=\frac{7}{8}$ and uses 0.5 and their $S_{\infty}$ at any point |
|---|---|---|
| $160\left(\frac{7}{8}\right)^N < (0.5)$ or $\left(\frac{7}{8}\right)^N < \left(\frac{0.5}{160}\right)$ | dM1 | Attempt to isolate $+160\left(\frac{7}{8}\right)^N$ or $\left(\frac{7}{8}\right)^N$ |
| $N\log\left(\frac{7}{8}\right) < \log\left(\frac{0.5}{160}\right)$ | M1 | Uses law of logarithms to obtain equation or inequality of form $N\log\left(\frac{7}{8}\right) < \log\left(\frac{0.5}{\text{their } S_{\infty}}\right)$ or $N > \log_{0.875}\left(\frac{0.5}{\text{their } S_{\infty}}\right)$ |
| $N > \frac{\log(\frac{0.5}{160})}{\log(\frac{7}{8})} = 43.19823...$ **cso** $\Rightarrow N = 44$ | A1 cso | $N=44$ (Allow $N \geq 44$ but no $N > 44$) |
**Alternative: Trial & Improvement:**
| Attempts $160 - S_N$ or $S_N$ with at least one value for $N > 40$ | M1 | |
|---|---|---|
| Attempts $160 - S_N$ or $S_N$ with $N=43$ or $N=44$ | dM1 | |
| Evidence of examining $160 - S_N$ or $S_N$ for **both** $N=43$ **and** $N=44$ with **both** values correct to 2 DP; e.g. $160 - S_{43}$ = awrt 0.51 and $160 - S_{44}$ = awrt 0.45 or $S_{43}$ = awrt 159.49 and $S_{44}$ = awrt 159.55 | M1 | |
| $N = 44$ | A1 cso | Answer of $N=44$ only with no working scores no marks |
---2. The first term of a geometric series is 20 and the common ratio is $\frac { 7 } { 8 }$. The sum to infinity of the series is $S _ { \infty }$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $S _ { \infty }$
The sum to $N$ terms of the series is $S _ { N }$
\item Find, to 1 decimal place, the value of $S _ { 12 }$
\item Find the smallest value of $N$, for which $S _ { \infty } - S _ { N } < 0.5$\\
2. The first term of a geometric series is 20 and the common ratio is $\frac { 7 } { 8 }$. The sum to infinity\\
of the series is $S _ { \infty }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2018 Q2 [8]}}