| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Form and solve quadratic in parameter |
| Difficulty | Standard +0.3 This is a standard geometric series question requiring students to use the property that consecutive terms have a constant ratio, leading to a quadratic equation. The algebra is straightforward, and all steps (forming equation, solving quadratic, finding ratio, applying sum formula) are routine C2 techniques with no novel insight required. Slightly easier than average due to heavy scaffolding in parts (a) and (b). |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a=4p\), \(ar=(3p+15)\), \(ar^2=5p+20\) | B1 | Correct statement of all three terms |
| \(\frac{5p+20}{3p+15}=\frac{3p+15}{4p}\) or \(4p(5p+20)=(3p+15)^2\) | M1 | Valid attempt to eliminate \(a\) and \(r\), obtain equation in \(p\) only |
| \((3p+15)^2=9p^2+90p+225\) | M1 | Correct expansion |
| \(11p^2-10p-225=0\) \(*\) | A1* | No incorrect work seen; printed answer obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((p-5)(11p+45)\) so \(p=\) | M1 | Attempt to solve by factorisation, completing square or formula |
| \(p=5\) only (rejecting \(-45/11\)) | A1 | 5 only; \(-45/11\) seen and rejected or \((11p+45)\) seen and \(p>0\) stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{3\times5+15}{4\times5}\) or \(\frac{5\times5+20}{3\times5+15}\) | M1 | Substitutes \(p=5\) and attempts ratio (correct way up) |
| \(r=\frac{3}{2}\) | A1 | 1.5 or any equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(S_{10}=\frac{20\!\left(1-\left(\frac{3}{2}\right)^{10}\right)}{1-\frac{3}{2}}\) | M1 A1ft | M1: correct formula with \(n=10\); A1ft: correct expression ft on their \(r\); must have \(a=20\) and power \(=10\) |
| \(=2267\) | A1 | Accept awrt 2267 |
## Question 5:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a=4p$, $ar=(3p+15)$, $ar^2=5p+20$ | B1 | Correct statement of all three terms |
| $\frac{5p+20}{3p+15}=\frac{3p+15}{4p}$ or $4p(5p+20)=(3p+15)^2$ | M1 | Valid attempt to eliminate $a$ and $r$, obtain equation in $p$ only |
| $(3p+15)^2=9p^2+90p+225$ | M1 | Correct expansion |
| $11p^2-10p-225=0$ $*$ | A1* | No incorrect work seen; printed answer obtained |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(p-5)(11p+45)$ so $p=$ | M1 | Attempt to solve by factorisation, completing square or formula |
| $p=5$ only (rejecting $-45/11$) | A1 | 5 only; $-45/11$ seen and rejected or $(11p+45)$ seen and $p>0$ stated |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{3\times5+15}{4\times5}$ or $\frac{5\times5+20}{3\times5+15}$ | M1 | Substitutes $p=5$ and attempts ratio (correct way up) |
| $r=\frac{3}{2}$ | A1 | 1.5 or any equivalent |
### Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $S_{10}=\frac{20\!\left(1-\left(\frac{3}{2}\right)^{10}\right)}{1-\frac{3}{2}}$ | M1 A1ft | M1: correct formula with $n=10$; A1ft: correct expression ft on their $r$; must have $a=20$ and power $=10$ |
| $=2267$ | A1 | Accept awrt 2267 |
---5. The first three terms of a geometric series are $4 p , ( 3 p + 15 )$ and ( $5 p + 20$ ) respectively, where $p$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $11 p ^ { 2 } - 10 p - 225 = 0$
\item Hence show that $p = 5$
\item Find the common ratio of this series.
\item Find the sum of the first ten terms of the series, giving your answer to the nearest integer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2013 Q5 [11]}}