## Question 8:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_n = a + ar + \ldots + ar^{n-1}$ and $rS_n = ar + ar^2 + \ldots + ar^n$ | M1 | Writes at least 3 correct terms of geometric series and multiplies by $r$; allow if 3 correct terms in both sequences and at least one "+" in both |
| $S_n - rS_n = a - ar^n$ or $rS_n - S_n = ar^n - a$ | A1(M1 on EPEN) | Both $S_n$ and $rS_n$ have correct first and last terms and at least one other correct term, no incorrect terms; both sides unsimplified |
| $(1-r)S_n = a(1-r^n) \Rightarrow S_n = \dfrac{a(1-r^n)}{1-r}$ * | A1* | Factorise both sides and divide by $1-r$; allow $S_n = \dfrac{a(1-r^n)}{(1-r)}$ but not $S_n = \dfrac{a(r^n-1)}{(r-1)}$ unless followed by correct version |

**Special case:** If terms listed rather than added and working otherwise correct, score 110.

**Alternative (a):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1-r)S_n = (1-r)(a + ar + \ldots + ar^{n-1})$ or $S_n = \dfrac{(1-r)(a+ar+\ldots+ar^{n-1})}{(1-r)}$ | M1 | Writes at least 3 correct terms and multiplies both sides by $1-r$ or multiplies RHS by $\dfrac{1-r}{1-r}$ |
| $(1-r)S_n = a - ar^n$ or $S_n = \dfrac{a - ar^n}{1-r}$ | A1(M1 on EPEN) | Correct first and last terms, at least one other correct term, no incorrect terms; RHS must be seen unfactorised unless $a$ was factored earlier |
| $(1-r)S_n = a - ar^n = a(1-r^n) \Rightarrow S_n = \dfrac{a(1-r^n)}{1-r}$ * | A1* | Should be as printed; allow $\dfrac{a(1-r^n)}{(1-r)}$ but not $\dfrac{a(r^n-1)}{(r-1)}$ unless followed by correct version |

**Proof by induction:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1 \Rightarrow S_1 = \dfrac{a(1-r^1)}{1-r} = a$; assume true for $n=k$ so $S_k = \dfrac{a(1-r^k)}{1-r}$ | M1 | Shows true for $n=1$, assumes true for $n=k$, adds $(k+1)$th term |
| $S_{k+1} = \dfrac{a(1-r^k)}{1-r} + ar^k = \dfrac{1-ar^k+ar^k-ar^{k+1}}{1-r} = \dfrac{a-ar^{k+1}}{1-r} = \dfrac{a(1-r^{k+1})}{1-r}$ | A1(M1 on EPEN) | Finds common denominator, obtains $\dfrac{a-ar^{k+1}}{1-r}$ using correct algebra |
| Fully correct proof reaching $\dfrac{a(1-r^{k+1})}{1-r}$ with all steps and conclusion | A1 | All steps shown with conclusion |

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### Parts (b) and (c) — marked together:

**Part (b):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r^3 = -\dfrac{20.48}{320} \Rightarrow r = \sqrt[3]{-\dfrac{20.48}{320}}$ | M1 | Correct strategy for $r$: dividing the 2 given terms either way round and attempting cube root |
| $r = -0.4$ | A1 | Correct value only; allow equivalents e.g. $-2/5$; correct answer scores both marks |

**Part (c):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = -0.4 \Rightarrow a = \dfrac{-320}{-0.4} = 800$ or $r = -0.4 \Rightarrow a = \dfrac{512}{25} \div \left(-\dfrac{2}{5}\right)^4 = 800$ | M1 | Correct attempt at first term using $\pm$ their $r$ and the $-320$ or $\dfrac{512}{25}$; using e.g. $ar = -320$ or $ar^4 = \dfrac{512}{25}$; **not** $ar^2 = -320$ or $ar^5 = \dfrac{512}{25}$ |
| $S_{13} = \dfrac{\text{"800"}(1 - \text{"}-0.4\text{"}^{13})}{1 - \text{"}-0.4\text{"}}$ | M1 | Correct attempt at sum using their $a$, their $r$, and $n=13$; must be fully correct attempt; note $\dfrac{800(1+0.4^{13})}{1+0.4}$ is equivalent to $\dfrac{800(1-(-0.4)^{13})}{1-(-0.4)}$ |
| $= 571.43$ | A1 | Correct value; note $S_\infty$ is also 571.43 so working must be seen; correct answer only scores no marks |

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