| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| 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 with standard techniques: finding the common ratio from two terms (part a is guided), then computing S_∞ and S_10 using formulas. The arithmetic is slightly tedious but requires no problem-solving insight, 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 | Mark | Guidance |
| \(125 \times \left(\frac{2}{5}\right)^3 = \ldots\) or \(8 \times \left(\frac{2}{5}\right)^{-3} = \ldots\) or \(r^3 = \frac{8}{125} \Rightarrow r = \ldots\) | M1 | Attempts to verify \(r = \frac{2}{5}\) using fourth term and \(r\) to reach seventh term or vice versa. Alternatively obtains correct equation for \(r^{\pm3}\) and proceeds to find \(r\). Accept \((r=)\sqrt[3]{\frac{8}{125}}\) |
| \(125 \times \left(\frac{2}{5}\right)^3 = 8 \Rightarrow r = \frac{2}{5}\) or \(r = \sqrt[3]{\frac{8}{125}} = \frac{2}{5}\) | A1* | Concludes \(r = \frac{2}{5}\) (requires a tick). Minimum acceptable: \(r = \sqrt[3]{\frac{8}{125}} = \frac{2}{5}\) or \(r^3 = \frac{8}{125} \Rightarrow r = \frac{2}{5}\). Ignore dubious working. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = \frac{125}{\left(\frac{2}{5}\right)^3}\) or \(a = \frac{8}{\left(\frac{2}{5}\right)^6}\) \(\left(= \frac{15625}{8} = 1953.125\right)\) | M1 | Correct method to find first term using \(r = \frac{2}{5}\). May be implied by \(\frac{15625}{8}\) or 1953.125 |
| \(S_\infty = \frac{\text{"}15625\text{"}/8}{1 - \frac{2}{5}}\) or \(S_{10} = \frac{\text{"}15625\text{"}/8\left(1-\left(\frac{2}{5}\right)^{10}\right)}{1-\frac{2}{5}}\) | M1 | Correct method to find \(S_\infty\) or \(S_{10}\) using their \(a\) and \(r = \frac{2}{5}\). If attempting \(S_{10}\) must use \(n=10\) |
| \(\pm(S_\infty - S_{10}) = \pm\left(\frac{78125}{24} - 3254.867\right) = \pm 0.341\) | dM1 | Both correct formulae for \(S_\infty\) and \(S_{10}\) and attempts \(\pm(S_\infty - S_{10})\). Dependent on previous method mark. |
| \(\pm(S_\infty - S_{10}) = \pm 0.341\) | A1 | awrt \(\pm 0.341\) |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $125 \times \left(\frac{2}{5}\right)^3 = \ldots$ or $8 \times \left(\frac{2}{5}\right)^{-3} = \ldots$ or $r^3 = \frac{8}{125} \Rightarrow r = \ldots$ | M1 | Attempts to verify $r = \frac{2}{5}$ using fourth term and $r$ to reach seventh term or vice versa. Alternatively obtains correct equation for $r^{\pm3}$ and proceeds to find $r$. Accept $(r=)\sqrt[3]{\frac{8}{125}}$ |
| $125 \times \left(\frac{2}{5}\right)^3 = 8 \Rightarrow r = \frac{2}{5}$ or $r = \sqrt[3]{\frac{8}{125}} = \frac{2}{5}$ | A1* | Concludes $r = \frac{2}{5}$ (requires a tick). Minimum acceptable: $r = \sqrt[3]{\frac{8}{125}} = \frac{2}{5}$ or $r^3 = \frac{8}{125} \Rightarrow r = \frac{2}{5}$. Ignore dubious working. |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \frac{125}{\left(\frac{2}{5}\right)^3}$ or $a = \frac{8}{\left(\frac{2}{5}\right)^6}$ $\left(= \frac{15625}{8} = 1953.125\right)$ | M1 | Correct method to find first term using $r = \frac{2}{5}$. May be implied by $\frac{15625}{8}$ or 1953.125 |
| $S_\infty = \frac{\text{"}15625\text{"}/8}{1 - \frac{2}{5}}$ or $S_{10} = \frac{\text{"}15625\text{"}/8\left(1-\left(\frac{2}{5}\right)^{10}\right)}{1-\frac{2}{5}}$ | M1 | Correct method to find $S_\infty$ or $S_{10}$ using their $a$ and $r = \frac{2}{5}$. If attempting $S_{10}$ must use $n=10$ |
| $\pm(S_\infty - S_{10}) = \pm\left(\frac{78125}{24} - 3254.867\right) = \pm 0.341$ | dM1 | Both correct formulae for $S_\infty$ and $S_{10}$ and attempts $\pm(S_\infty - S_{10})$. Dependent on previous method mark. |
| $\pm(S_\infty - S_{10}) = \pm 0.341$ | A1 | awrt $\pm 0.341$ |
---\begin{enumerate}
\item The 4th term of a geometric series is 125 and the 7th term is 8\\
(a) Show that the common ratio of this series is $\frac { 2 } { 5 }$\\
(b) Hence find, to 3 decimal places, the difference between the sum to infinity and the sum of the first 10 terms of this series.\\
\end{enumerate}
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel C12 2019 Q1 [6]}}