| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Total over time period |
| Difficulty | Moderate -0.3 This is a straightforward geometric series application with scaffolding. Part (a) is trivial pattern continuation (1 mark). Part (b)(i) requires recognizing a GP and applying the sum formula with n=120, which is standard A-level technique. Part (b)(ii) asks for a simple real-world explanation (interest rates may change). The question guides students through the setup and requires only routine application of series formulas, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks |
|---|---|
| 7(a) | Obtains |
| Answer | Marks | Guidance |
|---|---|---|
| or better seen | 2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(b)(i) | Models the total as the sum to n |
| Answer | Marks | Guidance |
|---|---|---|
| or 505.53 | 3.3 | M1 |
Total in account = £ 6 787
| Answer | Marks | Guidance |
|---|---|---|
| x=1 | 3.3 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| £6 787, £6 787.15 or £6 787.16 | 3.2a | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(b)(ii) | Makes a reasonable comment in |
| Answer | Marks | Guidance |
|---|---|---|
| payments. | 3.5b | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question 7 Total | 5 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Obtains
50×1.002 3 +50×1.002 2 +50×1.002
with
( )
50×1.002 2 +50×1.002+50 ×1.002
or better seen | 2.1 | R1 | ( )
T = 50×1.002 2 +50×1.002+50 ×1.002
3
=50×1.002 3 +50×1.002 2 +50×1.002
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b)(i) ---
7(b)(i) | Models the total as the sum to n
terms of a geometric sequence
Evidence for this could include
at least two of a, r or n
substituted into sum formula
a = 50.1, r = 1.002,
n = 120
Condone a = 50 or
n = 10 or 119
PI by AWRT 6724, 6737, 6801
or 505.53 | 3.3 | M1 | ( )
50.1 1−1.002 120
T =
120 1−1.002
=6787.1595...
Total in account = £ 6 787
Forms the correct expression for
the correct total
( )
50.1 1−1.002 120
( =)
T or
120 1−1.002
120
∑50×1.002x
x=1 | 3.3 | A1
Obtains
£6 787, £6 787.15 or £6 787.16 | 3.2a | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b)(ii) ---
7(b)(ii) | Makes a reasonable comment in
context. For example:
The interest rate is unlikely to
remain fixed.
Or
May have needed to withdraw
some amount.
Or
May change the monthly
payments. | 3.5b | E1 | The interest rate is unlikely to remain
fixed for the whole 10 years
Subtotal | 1
Question 7 Total | 5
Q | Marking instructions | AO | Marks | Typical solutionOn the first day of each month, Kate pays £50 into a savings account.
Interest is paid on the total amount in the account on the last day of each month.
The interest rate is 0.2%
At the end of the $n$th month, the total amount of money in Kate's savings account is £$T_n$
Kate correctly calculates $T_1$ and $T_2$ as shown below:
$T_1 = 50 \times 1.002 = 50.10$
$T_2 = (T_1 + 50) \times 1.002$
$= ((50 \times 1.002) + 50) \times 1.002$
$= 50 \times 1.002^2 + 50 \times 1.002$
$\approx 100.30$
\begin{enumerate}[label=(\alph*)]
\item Show that $T_3$ is given by
$$T_3 = 50 \times 1.002^3 + 50 \times 1.002^2 + 50 \times 1.002$$
[1 mark]
\item Kate uses her method to correctly calculate how much money she can expect to have in her savings account at the end of 10 years.
\begin{enumerate}[label=(\roman*)]
\item Find the amount of money Kate expects to have in her savings account at the end of 10 years.
[3 marks]
\item The amount of money in Kate's savings account at the end of 10 years may not be the amount she has correctly calculated.
Explain why.
[1 mark]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2024 Q7 [5]}}