| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Largest or extreme value of sum |
| Difficulty | Moderate -0.3 This is a straightforward arithmetic sequence question testing standard formulas. Part (a) is direct substitution into S_n = n/2[2a + (n-1)d]. Part (b) requires solving simultaneous equations, which is routine. Part (c) asks for explanation of a maximum, requiring recognition that terms change sign, but this is a standard observation. All parts use well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks |
|---|---|
| 5(a) | Uses S =260 for arithmetic |
| Answer | Marks | Guidance |
|---|---|---|
| PI by 8 ( 2a+15d )=260 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| answer. | 2.1 | R1 |
| Answer | Marks |
|---|---|
| 5(b) | Forms a second equation in a |
| Answer | Marks | Guidance |
|---|---|---|
| or d | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains correct a and d | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| their a or d is correct. | 1.1b | A1F |
| Answer | Marks |
|---|---|
| 5(c) | Explains that values of U are |
| Answer | Marks | Guidance |
|---|---|---|
| CSO | 2.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| CSO | 2.1 | R1 |
| Total | 7 | |
| Q | Marking Instructions | AO |
Question 5:
--- 5(a) ---
5(a) | Uses S =260 for arithmetic
n
sequence with n=16 to form a
correct equation
PI by 8 ( 2a+15d )=260 | 1.1a | M1 | 16
( 2a+( 16−1 ) d ) =260
2
8 ( 2a+15d )=260
2 ( 2a+15d )=65
4a+30d =65
Completes rigorous argument
with correct algebraic
manipulation to show required
result
Must see at least one line of
simplification after
8 ( 2a+15d )=260 before given
answer. | 2.1 | R1
--- 5(b) ---
5(b) | Forms a second equation in a
and d using S =315and
60
solves simultaneously to find a
or d | 3.1a | M1 | 30 ( 2a+59d )=315
20a+590d =105
a =20
d =−0.5
41
S = ( 2×20−40×0.5 )=410
41 2
Obtains correct a and d | 1.1b | A1
Uses their a and d to obtain their
value ofS =41a+820d
41
Follow through provided one of
their a or d is correct. | 1.1b | A1F
--- 5(c) ---
5(c) | Explains that values of U are
n
positive n < 41
Or
Explains that values of U are
n
negative for n>41
Or
Uses quadratic manipulation or
differentiation of formula for S
n
to obtain n = 40.5
CSO | 2.4 | M1 | The terms before the 41st term are
all positive. The terms after the 41st
term are all negative so the sum of
the first 41 terms must be a
maximum value.
Completes a valid argument
explaining all terms positive
before 41 and negative after 41
Or
Completes argument linking
40.5 with the sum to 40 terms
and the sum to 41 terms.
CSO | 2.1 | R1
Total | 7
Q | Marking Instructions | AO | Marks | Typical SolutionAn arithmetic sequence has first term $a$ and common difference $d$.
The sum of the first 16 terms of the sequence is 260
\begin{enumerate}[label=(\alph*)]
\item Show that $4a + 30d = 65$ [2 marks]
\item Given that the sum of the first 60 terms is 315, find the sum of the first 41 terms. [3 marks]
\item $S_n$ is the sum of the first $n$ terms of the sequence.
Explain why the value you found in part (b) is the maximum value of $S_n$ [2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2019 Q5 [7]}}