# Question 10:

## Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $d = \frac{37-15}{11} (= 2)$ | M1 | Attempts to find common difference. Allow $d = \frac{37-15}{k}$ where $k = 11$ or $12$. Must use 37 and 15. |
| $u_5 = 15 + 4 \times \text{"2"} = \ldots$ | M1 | Use their common difference to find distance on Sunday of week 5. Must be correct formula. May list first 5 terms (look for 5 terms). |
| Week 5 Sunday run $= 23$ km | A1 | |

## Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $r^{11} = \frac{37}{15} \Rightarrow r = \sqrt[11]{\frac{37}{15}} = \ldots$ | M1 | Correct method to find $r$. Award for $r^{11} = \frac{37}{15}$ and proceeding to find $r$. May be implied by awrt 1.09. Allow if $r^{11} = \frac{37}{15}$ followed by stating awrt 8.6% increase if no value given for $r$. |
| $u_5 = 15 \times \text{"1.0855..."}^4 = \ldots$ | M1 | Uses their $r$ to find distance on Sunday of week 5. Must be correct formula. May list first 5 terms (look for 5 terms). |
| Week 5 Sunday run $=$ awrt $20.8$ km | A1 | |

## Part (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $S_{12} = \frac{12}{2}[2\times15 + (12-1)\times\text{"2"}]$ or $\frac{12}{2}(15+37)$ $(= 312)$ | M1 | Attempts total distance over 12 weeks using model $A$ with $n=12$ and either $a=15$ and their $d$, or $a=15$ and $l=37$. Formula must be correct with values in correct places. Listing requires all 12 terms. |
| $S_{12} = \frac{15(1-\text{"1.08554..."}^{12})}{1-\text{"1.08554..."}}$ $(= 294.185\ldots)$ | M1 | Attempts total distance over 12 weeks using model $B$ with $a=15$ and their $r$. Formula must be correct. Listing requires all 12 terms. |
| $\text{Difference} = \left(\frac{360-312}{12}\right)$ or $\left(\frac{360-294.185\ldots}{12}\right) = \ldots$ | dM1 | Attempts to find weekly Wednesday amount for **either** model A **or** model B. Look for $\frac{360 - \text{their total}}{12}$. May set up inequality $\text{total} + 12x \leqslant 360$. Dependent on previous method mark. Allow $=, <$ instead of $\leqslant$ or inequalities in wrong direction. |
| Either: Training model $A$: $x \leqslant 4$ km, Training model $B$: $x \leqslant 5.4\ldots$ km, with $=$ or $\leqslant$ | A1 | For identifying **either** 4 for model A **or** 5 for model B as critical values. May be part of inequality. Note: use of $r = 1.09$ gives 4.8… instead of 5.4… and will not be eligible for A mark if model A also incorrect. Allow $=, <$ instead of $\leqslant$. |
| Training model $A$: $x = 4$ km, Training model $B$: $x = 5$ km | A1cso | Must be clear correct values are linked to correct models, but need not be labelled "model A" and "model B". |