| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Moderate -0.8 This is a straightforward geometric series question requiring only standard formula application: part (a) uses S_n formula with given values, part (b) applies S_∞ = a/(1-r) directly, and part (c) finds individual terms using ar^n. All three parts are routine calculations with no problem-solving or conceptual challenges, making it easier than average for A-level. |
| 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 |
| \(r = \frac{3}{4}\), \(S_4 = 175\) | ||
| Way 1: \(a\left(1-\left(\frac{3}{4}\right)^4\right)/(1-\frac{3}{4})\) or \(a(1-0.75^4)/(1-0.75)\) | M1 | Substituting \(r = \frac{3}{4}\) or 0.75 and \(n=4\) into formula for \(S\) |
| \(175 = \frac{a\left(1-\left(\frac{3}{4}\right)^4\right)}{1-\frac{3}{4}} \Rightarrow a = \frac{175(1-\frac{3}{4})}{\left(1-\left(\frac{3}{4}\right)^4\right)} \Rightarrow a = 64^*\) | A1* | Correct proof |
| [2] | ||
| Way 2: \(a + a\left(\frac{3}{4}\right) + a\left(\frac{3}{4}\right)^2 + a\left(\frac{3}{4}\right)^3\) | M1 | |
| \(\frac{175}{64}a = 175 \Rightarrow a = \frac{175}{\left(\frac{175}{64}\right)} \Rightarrow a = 64^*\) or \(2.734375a = 175 \Rightarrow a = 64\) | A1* | Correct proof |
| [2] | ||
| Way 3: \(\{S_4 = \} \frac{64\left(1-\left(\frac{3}{4}\right)^4\right)}{1-\frac{3}{4}}\) or \(\frac{64(1-0.75^4)}{1-0.75}\) | M1 | Applying formula for \(S_n\) with \(r = \frac{3}{4}\), \(n=4\) and \(a=64\) |
| \(= 175\) so \(a = 64^*\) | A1* | Obtains 175 with no errors and concludes \(a = 64^*\) |
| [2] | ||
| (b) \(\{S_\infty\} = \frac{64}{1-\frac{3}{4}}\); 256 | M1; A1cso | \(S_\infty = \frac{\text{(their } a\text{)}}{1-\frac{3}{4}}\) or \(\frac{64}{1-\frac{3}{4}}\); 256 |
| [2] | ||
| (c) Writes down either "\(64\left(\frac{3}{4}\right)^8\)" or awrt 6.4 or "\(-64\left(\frac{3}{4}\right)^8\)" or awrt 4.8, using \(a=64\) or their \(a\) | M1 | Can be implied |
| \(\{D = T_y - T_{10} = \} 64\left(\frac{3}{4}\right)^9 - 64\left(\frac{3}{4}\right)^8\) or awrt 4.8, using \(a=64\) or their \(a\) | Using \(a = 64\) (or their \(a\) found in part (a)) | |
| A correct expression for the difference \((i.e. \pm(T_y - T_{10}))\) using \(a=64\) or their \(a\) | dM1 | |
| \(= 64\left(\frac{3}{4}\right)\left(\frac{1}{4}\right) = 1.6018066...\) | A1 cao | 1.602 or \(-1.602\) |
| [3] |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = \frac{3}{4}$, $S_4 = 175$ | | |
| **Way 1:** $a\left(1-\left(\frac{3}{4}\right)^4\right)/(1-\frac{3}{4})$ or $a(1-0.75^4)/(1-0.75)$ | M1 | Substituting $r = \frac{3}{4}$ or 0.75 and $n=4$ into formula for $S$ |
| $175 = \frac{a\left(1-\left(\frac{3}{4}\right)^4\right)}{1-\frac{3}{4}} \Rightarrow a = \frac{175(1-\frac{3}{4})}{\left(1-\left(\frac{3}{4}\right)^4\right)} \Rightarrow a = 64^*$ | A1* | Correct proof |
| | [2] | |
| **Way 2:** $a + a\left(\frac{3}{4}\right) + a\left(\frac{3}{4}\right)^2 + a\left(\frac{3}{4}\right)^3$ | M1 | |
| $\frac{175}{64}a = 175 \Rightarrow a = \frac{175}{\left(\frac{175}{64}\right)} \Rightarrow a = 64^*$ or $2.734375a = 175 \Rightarrow a = 64$ | A1* | Correct proof |
| | [2] | |
| **Way 3:** $\{S_4 = \} \frac{64\left(1-\left(\frac{3}{4}\right)^4\right)}{1-\frac{3}{4}}$ or $\frac{64(1-0.75^4)}{1-0.75}$ | M1 | Applying formula for $S_n$ with $r = \frac{3}{4}$, $n=4$ and $a=64$ |
| $= 175$ so $a = 64^*$ | A1* | Obtains 175 with no errors and concludes $a = 64^*$ |
| | [2] | |
| **(b)** $\{S_\infty\} = \frac{64}{1-\frac{3}{4}}$; 256 | M1; A1cso | $S_\infty = \frac{\text{(their } a\text{)}}{1-\frac{3}{4}}$ or $\frac{64}{1-\frac{3}{4}}$; 256 |
| | [2] | |
| **(c)** Writes down either "$64\left(\frac{3}{4}\right)^8$" or awrt 6.4 or "$-64\left(\frac{3}{4}\right)^8$" or awrt 4.8, using $a=64$ or their $a$ | M1 | Can be implied |
| $\{D = T_y - T_{10} = \} 64\left(\frac{3}{4}\right)^9 - 64\left(\frac{3}{4}\right)^8$ or awrt 4.8, using $a=64$ or their $a$ | | Using $a = 64$ (or their $a$ found in part (a)) |
| A correct expression for the difference $(i.e. \pm(T_y - T_{10}))$ using $a=64$ or their $a$ | dM1 | |
| $= 64\left(\frac{3}{4}\right)\left(\frac{1}{4}\right) = 1.6018066...$ | A1 cao | 1.602 or $-1.602$ |
| | [3] | |
---\begin{enumerate}
\item A geometric series has first term $a$ and common ratio $r = \frac { 3 } { 4 }$
\end{enumerate}
The sum of the first 4 terms of this series is 175\\
(a) Show that $a = 64$\\
(b) Find the sum to infinity of the series.\\
(c) Find the difference between the 9th and 10th terms of the series. Give your answer to 3 decimal places.\\
\hfill \mbox{\textit{Edexcel C2 2016 Q1 [7]}}