| Exam Board | OCR MEI |
|---|---|
| Module | Further Extra Pure (Further Extra Pure) |
| Year | 2019 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Applied recurrence modeling |
| Difficulty | Standard +0.8 This is a multi-part recurrence relation problem requiring modeling a loan repayment scheme, solving a first-order linear recurrence relation, analyzing convergence conditions, and applying the solution to a specific numerical case. While the individual techniques (recurrence relations, geometric series) are standard Further Maths topics, the question requires sustained reasoning across multiple parts, translating between financial context and mathematical formalism, and careful algebraic manipulation. The convergence condition in part (c) requires insight into when the sequence becomes negative. This is more demanding than typical A-level questions but not exceptionally difficult for Further Maths students. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.10a General/particular solutions: of differential equations |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | æ a ö æ a ö |
| Answer | Marks |
|---|---|
| 100 | B1B1 |
| [2] | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | Trial solution of form L = kpn |
| Answer | Marks |
|---|---|
| 0 n è 1-aø 1-a | M1 |
| Answer | Marks |
|---|---|
| [5] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | Use of auxiliary equation |
| Answer | Marks |
|---|---|
| Must be C, a, b, n for final A1 | L = kan M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | Alternative Solution: |
| Answer | Marks |
|---|---|
| ë | 1.1 |
| 1.1 | At least L for M1. A1 if correct |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | b R |
| Answer | Marks |
|---|---|
| a/100 100 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | Allow their coefficient of an < 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (d) | (i) |
| Answer | Marks |
|---|---|
| Repayment takes 21 years | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 3.2a | For attempt at substitution into |
| Answer | Marks |
|---|---|
| n > 0.919 (3sf) B1 | Allow starting again |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (d) | (ii) |
| Answer | Marks |
|---|---|
| Total repayment is £62 746 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 3.2a | n´3000+L where n is their |
| Answer | Marks |
|---|---|
| (Accept rounding to nearest £) | (£62 746.25) |
Question 5:
5 | (a) | æ a ö æ a ö
L = L + L -R= 1+ L -R
ç ÷ ç ÷
n+1 n è100ø n è 100ø n
a
so a=1+ and b=-R
100 | B1B1
[2] | 3.3
3.3
5 | (b) | Trial solution of form L = kpn
n
gives p = a
b
Particular solution L =cÞc=
n 1-a
b
so general solution is L = kan+
n 1-a
æ b ö b
L =CÞL = C- an+
ç ÷
0 n è 1-aø 1-a | M1
A1
M1
A1ft
A1
cao
[5] | 1.1a
1.1
1.1
1.1
1.1 | Use of auxiliary equation
Allow L = their CS + PS as long
n
as PS is constant
Must be C, a, b, n for final A1 | L = kan M1A1
n
5 | (b) | Alternative Solution:
L = aL +b
n+1 n
L =CÞ L = aC+b, L = a2C+b(a+1)...
0 1 2
L = anC+(an-1b+....+ab+a)
n
b(1-an)
L = anC+
n 1-a
é æ b ö b ù
êÞ L = ç C- ÷ an+ ú
n è 1-aø 1-aû
ë | 1.1
1.1 | At least L for M1. A1 if correct
2
soi
GP formula
Must be C, a, b, n for final A1
If quoting solution must be fully
correct B5
M1A1
A1
M1A1
[5]
5 | (c) | b R
Terminates if C- <0ÞC- <0
1-a a/100
R aC
so >CÞR>
a/100 100 | M1
A1
[2] | 3.4
1.1 | Allow their coefficient of an < 0
AG; no slips in details of working
5 | (d) | (i) | L =37500-7500´1.08n
n
ln5
L <0Þn>
n ln1.08
Repayment takes 21 years | M1
A1
A1cao
[3] | 3.4
1.1
3.2a | For attempt at substitution into
their expression for
L =s´1.08n +t oe.
n
Must contain 1.08
For awrt 20.9 oe. Allow = or
(incorrect) inequality signs
SC Putting a = 0.08 gives
n > 0.919 (3sf) B1 | Allow starting again
and using GP formula
NB Must be algebraic
Alternative Solution:
L = aL +b
n+1 n
L =CÞ L = aC+b, L = a2C+b(a+1)...
0 1 2
L = anC+(an-1b+....+ab+a)
n
b(1-an)
L = anC+
n 1-a
é æ b ö b ù
êÞ L = ç C- ÷ an+ ú
n è 1-aø 1-aû
ë
At least L for M1. A1 if correct
2
soi
GP formula
Must be C, a, b, n for final A1
If quoting solution must be fully
correct B5
1.1
1.1
5 | (d) | (ii) | L =-253.75
21
21´3000-253.75
Total repayment is £62 746 | B1
M1
A1cao
[3] | 3.4
1.1
3.2a | n´3000+L where n is their
n
5d(i)
(Accept rounding to nearest £) | (£62 746.25)A financial institution models the repayment of a loan to a client in the following way.
\begin{itemize}
\item An amount, $£C$, is loaned to the client at the start of the repayment period.
\item The amount owed $n$ years after the start of the repayment period is $£L_n$, so that $L_0 = C$.
\item At the end of each year, interest of $\alpha\%$ ($\alpha > 0$) of the amount owed at the start of that year is added to the amount owed.
\item Immediately after interest has been added to the amount owed a repayment of $£R$ is made by the client.
\item Once $L_n$ becomes negative the repayment is finished and the overpayment is refunded to the client.
\end{itemize}
\begin{enumerate}[label=(\alph*)]
\item Show that during the repayment period, $L_{n+1} = aL_n + b$, giving $a$ and $b$ in terms of $\alpha$ and $R$. [2]
\item Find the solution of the recurrence relation $L_{n+1} = aL_n + b$ with $L_0 = C$, giving your solution in terms of $a$, $b$, $C$ and $n$. [5]
\item Deduce from parts (a) and (b) that, for the repayment scheme to terminate, $R > \frac{\alpha C}{100}$. [2]
\end{enumerate}
A client takes out a £30000 loan at 8% interest and agrees to repay £3000 at the end of each year.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item \begin{enumerate}[label=(\roman*)]
\item Use an algebraic method to find the number of years it will take for the loan to be repaid. [3]
\item Taking into account the refund of overpayment, find the total amount that the client repays over the lifetime of the loan. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Extra Pure 2019 Q5 [15]}}