| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| 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 structured multi-part question on geometric sequences that guides students through forming and solving a quadratic equation. Part (a) requires using the property that consecutive terms have a constant ratio (leading to a quadratic), parts (b)-(c) involve routine algebraic manipulation and applying standard GP formulas. While it requires multiple steps, each step is clearly signposted and uses standard C2 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks |
|---|---|
| (a) \(a=7k-5\), \(ar=5k-7\) and \(ar^{2}=2k+10\) | B1 |
| \((So\ r=)\) \(\frac{5k-7}{7k-5}=\frac{2k+10}{5k-7}\) or \((7k-5)(2k+10)=(5k-7)^{2}\) or equivalent | M1 |
| See \((5k-7)^{2}=25k^{2}-70k+49\) | M1 |
| \(14k^{2}+60k-50=25k^{2}-70k+49\rightarrow 11k^{2}-130k+99=0\) * | A1cso * |
| (b) \((k-11)(11k-9)\) so \(k=\) | M1 |
| \(k=\frac{9}{11}\) only* (after rejecting 11) | A1* |
| N.B. Special case \(k=\frac{9}{11}\) can be verified in (b) (1 mark only) | M1A0 |
| (c) \(a=\frac{8}{11}\) | B1 |
| \(\frac{5x\frac{9}{11}-7}{7x\frac{9}{11}-5}\) or \(\frac{2x\frac{9}{11}+10}{5x\frac{9}{11}-7}\) so \(r=-4\) | B1 |
| (i) Fourth term = \(ar^{3}=-\frac{512}{11}\) | M1A1 |
| (ii) \(S_{10}=\frac{a(1-r^{10})}{(1-r)}=\frac{\frac{8}{11}(1-(-4)^{10})}{(1-(-4))}=-152520\) | M1A1 |
| (a) $a=7k-5$, $ar=5k-7$ and $ar^{2}=2k+10$ | B1 |
|---|---|
| $(So\ r=)$ $\frac{5k-7}{7k-5}=\frac{2k+10}{5k-7}$ or $(7k-5)(2k+10)=(5k-7)^{2}$ or equivalent | M1 |
| See $(5k-7)^{2}=25k^{2}-70k+49$ | M1 |
| $14k^{2}+60k-50=25k^{2}-70k+49\rightarrow 11k^{2}-130k+99=0$ * | A1cso * |
| (b) $(k-11)(11k-9)$ so $k=$ | M1 |
| $k=\frac{9}{11}$ only* (after rejecting 11) | A1* |
| **N.B. Special case** $k=\frac{9}{11}$ can be verified in (b) (1 mark only) | M1A0 |
| (c) $a=\frac{8}{11}$ | B1 |
| $\frac{5x\frac{9}{11}-7}{7x\frac{9}{11}-5}$ or $\frac{2x\frac{9}{11}+10}{5x\frac{9}{11}-7}$ so $r=-4$ | B1 |
| (i) Fourth term = $ar^{3}=-\frac{512}{11}$ | M1A1 |
| (ii) $S_{10}=\frac{a(1-r^{10})}{(1-r)}=\frac{\frac{8}{11}(1-(-4)^{10})}{(1-(-4))}=-152520$ | M1A1 |
**Notes:**
- **(a) Mark parts (a) and (b) together**
- B1: Correct statement (needs all three terms) – this may be omitted and implied by correct statement in $k$ only, as candidates may use geometric mean, or may use ratio of terms being equal and give a correct line 2 without line 1. (This would earn B1M1 immediately)
- M1: Valid Attempt to eliminate $a$ and $r$ to obtain equation in $k$ only
- M1: Correct expansion of $(5k-7)^{2}=25k^{2}-70k+49$ - may have four terms $(5k-7)^{2}=25k^{2}-35k-35k+49$
- A1cso: No incorrect work seen. The printed answer is obtained including "=0".
- **(b) M1:** Attempt to solve quadratic by usual methods (factorisation, completion of square or formula – see notes at start of mark scheme) or see 9/11 substituted and given as "=0" for M1A0
- **A1\*:** 9/11 and 11 should be seen and rejected. Accept 9/11 underlined or $k=$ 9/11 written on following line.
- **Alternatively** $(k-11)$ may be seen in the factorisation and a statement 'k not integer' given with $k=$ 9/11 stated.
- **NB 0.81** with no negative sign in numerator can be allowed M1A0 but if the incorrect numerical expression appears on its own with no formula then M0A0 unless if incorrect numerical expression appears on its own with no formula then M0A0 but if the incorrect numerical expression appears on its own
- **(c) Mark parts (i) and (ii) together**
- B1: $a=\frac{8}{11}$ or any equivalent (If not stated explicitly or used in formula may be implied by correct answer to (ii))
- B1:Substitutes $k=9/11$ completely and obtain $r=-4$ (If not stated explicitly, may be implied by correct answer to (i) or (iii))
- A1: Use of correct formula with $n=10$, $a$ and/or $r$ may still be in terms of $k$. May assume $r=k$
- **(i) M1:** Use of correct formula with $n=10$, $a$ and/or $r$ may still be in terms of $k$. May assume $r=k$; **A1:** $-152520$ cao
- **NB Correct formula with negative sign in numerator followed by the incorrect $(8/11)(1-4)^{(10)})$ usually found equal to 152520.2909 with no negative sign can be allowed M1A0 but if the incorrect numerical expression appears on its own9. The first three terms of a geometric sequence are
$$7 k - 5,5 k - 7,2 k + 10$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $11 k ^ { 2 } - 130 k + 99 = 0$
Given that $k$ is not an integer,
\item show that $k = \frac { 9 } { 11 }$
For this value of $k$,
\item \begin{enumerate}[label=(\roman*)]
\item evaluate the fourth term of the sequence, giving your answer as an exact fraction,
\item evaluate the sum of the first ten terms of the sequence.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2017 Q9 [12]}}