| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find common ratio from terms |
| Difficulty | Moderate -0.8 This is a straightforward geometric sequence question requiring basic formula manipulation (ar^n = term) to find r from two given terms, then simple substitution to find the first term and sum to infinity. The algebra is routine (solving r² = 1/16) and all techniques are standard C2 content with no problem-solving insight required. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(ar = 48\); \(ar^3 = 3\) | B1 |
| \(\Rightarrow 16r^2 = 1\) | M1 | Elimination of \(a\) OE |
| \(r^2 = \frac{1}{16} \Rightarrow r = -\frac{1}{4}\) or \(r = \frac{1}{4}\) | A1 | CSO AG Full valid completion. SC Clear explicit verification (max B2 out of 3.) |
| Total: 4 | ||
| (b)(i) | \(a = -192\) | B1 |
| (ii) | \(\frac{a}{1-r} = \frac{a}{1-(-\frac{1}{4})}\) | M1 |
| \(S_\infty = \frac{-768}{5} (= -153.6)\) | A1ft | Ft on candidate's value for \(a\). i.e. \(\frac{4}{5}a\) SC candidate uses \(r = 0.25\), gives \(a = 192\) and sum to infinity \(= 256\). (max. B0 M1A1) |
| Total: 7 |
(a) | $ar = 48$; $ar^3 = 3$ | B1 | For either. OE |
| $\Rightarrow 16r^2 = 1$ | M1 | Elimination of $a$ OE |
| $r^2 = \frac{1}{16} \Rightarrow r = -\frac{1}{4}$ or $r = \frac{1}{4}$ | A1 | CSO AG Full valid completion. SC Clear explicit verification (max B2 out of 3.) |
| | **Total: 4** | |
(b)(i) | $a = -192$ | B1 | 1 |
| | | |
(ii) | $\frac{a}{1-r} = \frac{a}{1-(-\frac{1}{4})}$ | M1 | $\frac{a}{1-r}$ used |
| | | |
| $S_\infty = \frac{-768}{5} (= -153.6)$ | A1ft | Ft on candidate's value for $a$. i.e. $\frac{4}{5}a$ SC candidate uses $r = 0.25$, gives $a = 192$ and sum to infinity $= 256$. (max. B0 M1A1) |
| | **Total: 7** | |
---5 The second term of a geometric series is 48 and the fourth term is 3 .
\begin{enumerate}[label=(\alph*)]
\item Show that one possible value for the common ratio, $r$, of the series is $- \frac { 1 } { 4 }$ and state the other value.
\item In the case when $r = - \frac { 1 } { 4 }$, find:
\begin{enumerate}[label=(\roman*)]
\item the first term;
\item the sum to infinity of the series.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2007 Q5 [7]}}