| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Standard summation formula application |
| Difficulty | Standard +0.3 This question tests telescoping series and recognizing that P(n) = n³. Part (a) is straightforward calculation, while part (b) requires taking a cube root. The conceptual insight that the difference of consecutive sums equals the last term is standard, and the arithmetic is routine, making this slightly easier than average. |
| Spec | 1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | Obtains one correct value | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains both correct values | 1.1b | B1 |
| Answer | Marks |
|---|---|
| 8(b) | Forms cubic equation replacing |
| Answer | Marks | Guidance |
|---|---|---|
| PI correct answer | 2.5 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| CSO | 1.1b | A1 |
| Total | 4 | |
| Q | Marking instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Obtains one correct value | 1.1b | B1 | ( )=27
P 3
( )=1000
P 10
Obtains both correct values | 1.1b | B1
--- 8(b) ---
8(b) | Forms cubic equation replacing
P ( n )=n3 (condone k3)
PI correct answer | 2.5 | M1 | n3 =1.25×108
n=500
Obtains 500
CSO | 1.1b | A1
Total | 4
Q | Marking instructions | AO | Mark | Typical solution$$P(n) = \sum_{k=0}^{n} k^3 - \sum_{k=0}^{n-1} k^3 \text{ where } n \text{ is a positive integer.}$$
\begin{enumerate}[label=(\alph*)]
\item Find P(3) and P(10) [2 marks]
\item Solve the equation $P(n) = 1.25 \times 10^8$ [2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2019 Q8 [4]}}