| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Sum to infinity from S_n ratio |
| Difficulty | Standard +0.3 This is a straightforward geometric series question requiring standard formula manipulation. Students must use S_n = a(r^n - 1)/(r - 1), set up equations from given conditions, and solve algebraically. While it involves multiple parts and some algebraic manipulation, it follows a predictable pattern with no novel insights required, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{a(r^3-1)}{r-1} = 10 \times \frac{a(r^2-1)}{r-1}\) | B1 M1 | |
| \(r^4 - 1 = 10(r^2-1)\) | A1 | |
| \(r^4 - 10r^2 + 9 = 0\) | M1 | |
| \((r^2-1)(r^2-9) = 0\) | M1 | |
| \(r^2 = 1, 9\) | M1 | |
| \(r = \pm 1, \pm 3\) | A1 | |
| \(r > 1 \therefore r = 3\) | A1 | |
| (b) \(\frac{a(3^3-1)}{3-1} = 26\) | M1 A1 | |
| \(a = \frac{26}{13} = 2\) | A1 | |
| (c) \(S_6 = \frac{2(3^6-1)}{3-1} = 728\) | M1 A1 | (11) |
(a) $\frac{a(r^3-1)}{r-1} = 10 \times \frac{a(r^2-1)}{r-1}$ | B1 M1 |
$r^4 - 1 = 10(r^2-1)$ | A1 |
$r^4 - 10r^2 + 9 = 0$ | M1 |
$(r^2-1)(r^2-9) = 0$ | M1 |
$r^2 = 1, 9$ | M1 |
$r = \pm 1, \pm 3$ | A1 |
$r > 1 \therefore r = 3$ | A1 |
(b) $\frac{a(3^3-1)}{3-1} = 26$ | M1 A1 |
$a = \frac{26}{13} = 2$ | A1 |
(c) $S_6 = \frac{2(3^6-1)}{3-1} = 728$ | M1 A1 | (11)8. A geometric series has first term $a$ and common ratio $r$ where $r > 1$.
The sum of the first $n$ terms of the series is denoted by $S _ { n }$.
Given that $S _ { 4 } = 10 \times S _ { 2 }$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $r$.
Given also that $S _ { 3 } = 26$,
\item find the value of $a$,
\item show that $S _ { 6 } = 728$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q8 [11]}}