## Question 8(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $u_k = 50 \times 0.8^{k-1}$ | B1 | State correct $50 \times 0.8^{k-1}$. Allow B1 even if subsequently becomes $40^{k-1}$. Could be implied by a later (in)equation eg $0.8^{k-1} < 0.003$. Must be seen correct numerically so stating $a=50$, $r=0.8$, $u_k = ar^{k-1}$ is not enough |
| $50 \times 0.8^{k-1} < 0.15$; $0.8^{k-1} < 0.003$; $\log 0.8^{k-1} < \log 0.003$ | M1 | Link to 0.15, rearrange and introduce logs or equiv. Allow any sign, equality or inequality. Allow no, or consistent, log base on both sides or $\log_{0.8}$ on RHS. If starting with $\log(50 \times 0.8^{k-1}) < \log 0.15$ then LHS must be correctly split to $\log 50 + \log 0.8^{k-1}$ for M1. M0 if solving $40^{k-1} < 0.15$. Allow M1 if using $50 \times 0.8^k$. M0 if using $S_k$ |
| $(k-1)\log 0.8 < \log 0.003$ | A1 | Obtain correct linear (in)equation. Could be $(k-1)\log 0.8 < \log 0.003$, $(k-1) < \log_{0.8} 0.003$ or $\log 50 + (k-1)\log 0.8 < \log 0.15$. Allow no brackets if implied by later work. Allow any linking sign, including $>$ |
| $k > 27.03$; $k = 28$ | A1 | Obtain $k=28$ (equality only). Must be equality in words or symbols ie $k=28$ or $k$ is 28, but A0 for $k \geq 28$ or $k$ is at least 28. Allow BOD if inequality sign not correct throughout as long correct final conclusion |
| | [4] | Answer only, or trial and improvement, eligible for first B1 only. Allow $n$ not $k$ throughout |

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## Question 8(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $ar = -3$ | B1 | State $ar = -3$. Any correct statement, including $a \times r^{(2-1)} = -3$ etc |
| $\frac{a}{1-r} = 4$ | B1 | State $\frac{a}{1-r} = 4$. Any correct statement, not involving $r^\infty$ (unless it becomes 0) |
| $-\frac{3}{r} = 4(1-r)$ | M1* | Attempt to eliminate either $a$ or $r$. Using valid algebra so M0 for eg $a = -3 - r$. Must be using $ar^k$ and $\frac{\pm a}{(\pm 1 \pm r)}$. Award as soon as equation in one variable is seen |
| $4r^2 - 4r - 3 = 0$ / $a^2 - 4a - 12 = 0$ | A1 | Obtain correct simplified quadratic. Any correct quadratic not involving fractions or brackets ie $4r^2 = 4r + 3$ gets A1 |
| $(2r-3)(2r+1)=0$ / $(a-6)(a+2)=0$ | M1d* | Attempt to solve 3 term quadratic |
| $r = -\frac{1}{2}$ | M1** | Identify $r = -\frac{1}{2}$ as only ratio with a minimally acceptable reason. M0 if no, or incorrect, reason given. Must have correct quadratic, correct factorisation and correct roots (if stated). If $r = -\frac{1}{2}$ is not explicitly identified then allow M1 when they use only this value to find $a$ (or later eliminate the other value). Could accept $r = -\frac{1}{2}$ as $r < 1$ or reject $r = \frac{3}{2}$ as $> 1$. Could reject $a = -2$ as $S_\infty$ is positive. Could refer to convergent/divergent series |
| $a = 6$ | A1 | Obtain $a = 6$ only. If solving quadratic in $a$, then both values of $a$ may be seen initially - A1 can only be awarded when $a=6$ is given as only solution |
| for sum to infinity $-1 < r < 1$ | A1d** | Convincing reason for $r = -\frac{1}{2}$ as the only possible ratio. Must refer to $|r| < 1$ or $-1 < r < 1$ oe in words. A0 if additional incorrect statement |
| | [8] | No credit for answer only unless both $r$ first found |

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