# Question 9:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t_5 = ar^{n-1} = 20 \times 0.9^{5-1} = 13.122$ | M1A1 | M1: Use of correct formula with $a=20$, $r=0.9$, $n=5$. Can be implied by correct answer. A1: $13.122$ or $\frac{6561}{500}$. Apply isw but just $13.1$ is A0 |

**(2 marks)**

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_8 = \frac{a(1-r^n)}{1-r} = \frac{20(1-0.9^8)}{1-0.9} = 113.9$ | M1A1 | M1: Use of correct formula with $a=20$, $r=0.9$, $n=8$. A1: $113.9$ **only** |

**(2 marks)**

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_\infty = \frac{20}{1-0.9} (= 200)$ | B1 | Correct $S_\infty$ which can be simplified or un-simplified |
| $200 - \frac{20(1-0.9^N)}{1-0.9} < 0.04$ | M1A1 | M1: Attempts $S_\infty - S_N < 0.04$ (allow $n$ for $N$) using $a=20$ and $r=0.9$. A1: Correct inequality in any form in terms of $N$ or $n$ only |
| $0.9^N < 0.0002$ | A1* | Reaches the printed answer with intermediate working and with **no errors or incorrect statements** |

**(4 marks)**

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(N >) \frac{\log 0.0002}{\log 0.9} \Rightarrow N = 81$ | M1A1 | M1: Correct attempt to find $N$ ignoring what they use for "$>$". Look for $(N =)\frac{\log 0.0002}{\log 0.9}$ or $(N =)\log_{0.9} 0.0002$. May be implied by awrt 81. A1: 81 only. Accept $81$ only or $N/n = 81$ but not $N/n > 81$ |

**(2 marks) — Total 10**

---