## Question 7:

### Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_{20} = 5 + 19 \times 3$ | M1 | Must be using correct formula with $a=5$ and $d=3$. Could use $u_n = 3n+2$ |
| $= 62$ | A1 | If listing terms then need to indicate that 62 is the required answer |

### Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{20} = \frac{20}{2}(10+57)$, $S_9 = \frac{9}{2}(10+24)$ | M1 | Must be using correct formula with $a=5$ and $d=3$. Use of formula must be explicit |
| $\frac{20}{2}(10+57) - \frac{9}{2}(10+24)$ | M1 | Attempt $S_{20}-S_9$ where both summations have been shown explicitly |
| $= 670 - 153 = 517$ **AG** | A1 | AG so detail is required - only award A1 if both unsimplified sums are seen, as well as both evaluated sums. SR: Allow B1 if only $670-153=517$ seen |
| OR: $u_{10} = 5+9\times3=32$, $S = \frac{11}{2}(32+62)$ | M1, M1 | Must be shown explicitly; must have $n=11$ |
| $= 517$ **AG** | A1 | Detail reqd - award M0M1A0 if no evidence for $u_{10}=32$ |

### Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{2N} = \frac{2N}{2}(10+3(2N-1))$ | B1 | Or equiv from $\frac{1}{2}n(a+l)$, or from $\Sigma(3n+2)$ |
| $S_{N-1} = \frac{N-1}{2}(10+3(N-2))$ | B1 | Or $S_N - u_N$ soi |
| $N(6N+7) - \frac{N-1}{2}(3N+4) = 2750$ | M1* | Subtract attempt at $S_{N-1}$ from $S_{2N}$, equate to 2750. Allow sign errors from lack of essential brackets |
| $9N^2 + 13N - 5496 = 0$ | A1 | Rearrange to obtain $9N^2+13N-5496 = 0$ |
| $(9N+229)(N-24) = 0$ | M1d* | Attempt to solve 3-term quadratic to obtain at least the positive root |
| $N = 24$ | A1 | Obtain $N=24$ only CWO. No need to consider negative root, but A0 if found but not discarded |

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