## Question 4:

**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_1 = 6 \Rightarrow u_2 = 6k-5$; $u_2 = 6k-5 \Rightarrow u_3 = k(6k-5)-5$; $\Rightarrow k(6k-5)-5 = -1$ | M1 | Correct application of recurrence relation using $u_1=6$ to find $u_2$ then $u_3$ in terms of $k$, sets $u_3 = -1$; condone missing brackets if intention clear |
| $\Rightarrow 6k^2 - 5k - 4 = 0$ | A1* | Obtains printed answer with no errors including "$= 0$"; given answer so do not condone slips unless recovered before final answer |

**Part (b)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = \frac{4}{3}$ | B1 | Deduces correct value of $k$; allow $1\frac{1}{3}$ or $1.\dot{3}$ but not 1.333; if both roots offered score B0 unless $k=\frac{4}{3}$ clearly intended |

**Part (b)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = \frac{4}{3} \Rightarrow u_2 = \frac{4}{3} \times 6 - 5 \Rightarrow \sum_{r=1}^{3} u_r = 6 + \frac{4}{3} \times 6 - 5 - 1$ | M1 | Correct method using their $k$ value |
| $\sum_{r=1}^{3} u_r = 8$ | A1 | Correct answer |

# Question 4(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts second term e.g. $6 + "\frac{4}{3}" \times 6 - 5 - 1$ | M1 | Must use $(their\ k) \times 6 - 5$ then add 6 and $-1$; if using $u_1$ and $u_3$ they must be as given; attempt at second term may be implied by value |
| $\sum_{r=1}^{3} u_r = 8$ | A1 | Correct value of 8 and no other values unless rejected; correct answer with no working scores both marks; allow recovery from inexact value from part (i) e.g. 1.333 |

---