AS June 2025 Q3

EdexcelCurrent spec10 marksRecurrence Relations

3. A loan of £180 000 is taken out to buy a house.

The monthly interest rate on the loan is 0.15%

The interest is added to the balance of the loan at the end of each month.

To repay the loan, £900 is repaid at the end of each month, immediately after the interest has been added.

Let \(B_n\) thousands of pounds be the balance of the loan at the end of month \(n\) after the interest has been added and the £900 repaid.

(a) Explain, in the context of the problem, why the balance of the loan, \(B_n\), can be modelled by the recurrence relation\[B_n = 1.0015B_{n-1} - 0.9 \qquad B_0 = 180 \qquad n \in \mathbb{Z}^+\] (2)

(b) State an assumption that must be made for this model to be valid. (1)

(d) Hence determine the time it will take to repay the loan. Give your answer in years and months to the nearest month. (2)

Mark scheme (a)

Scheme	Marks	AO
The loan amount is £180 000 before any interest is added or payments made so \(B_0 = 180\) as the units are 1000’s Interest is added at 0.15% so the monthly balance is multiplied by 100.15% = 1.0015 to give \(1.0015B_{n-1}\) as the new balance After the interest has been added, £900 is paid off which is 0.9 in thousands of pounds so \(B_n = 1.0015B_{n-1} - 0.9\)	B1 B1	2.4 3.3
	(2)

Notes

B1: For explaining 2 of the 3 aspects as above. Allow attempts that convey the right idea even if not precisely described.

B1: All 3 aspects explained with sufficient detail shown. Must see the 1.0015 explained, not just “the 1.0015 is the 0.15%” or such.

Mark scheme (b)

Scheme	Marks	AO
E.g. The interest rate stays the same The monthly repayments stay the same	B1	3.5b
	(1)

Notes

B1: See scheme for answers. Must refer to the model, do not accept answer about “rounding” values.

Mark scheme (c)

Scheme	Marks	AO
A complete method to solve the recurrence relation using \(B_n = \text{CF} + \text{PS} = a(1.0015)^n + b\)	M1	3.1a
\(\text{PS} = b \;\Rightarrow\; b = 1.0015b - 0.9\) leading to \(b = \ldots\)	M1	1.1b
\(b = 600\)	A1	1.1b
Uses \(B_0 = 180\) and their value for \(b\) to find the value of \(a\) \(180 = a(1.0015)^0 + 600\) \(a = \ldots(-420)\)	M1	1.1b
\(B_n = 600 - 420(1.0015)^n \quad (n \geqslant 0)\)	A1	1.1b
	(5)

Notes

M1: A complete method to solve the recurrence relation using \(B_n = \text{CF} + \text{PS} = a(1.0015)^n + b\)

M1: Uses \(\text{PS} = b \Rightarrow b = 1.0015b - 0.9\) to find a value for \(b\) (corrected from the printed mark scheme: the printed note has \(b = 1.0015b - 900\), but \(B_n\) is in thousands of pounds, so the constant is 0.9 as in the scheme)

A1: \(b = 600\)

M1: Uses \(B_0\) and their value for \(b\) to find a value for \(a\)

A1: Fully correctly defined sequence \(B_n = 600 - 420(1.0015)^n \quad (n \geqslant 0)\)

Mark scheme (d)

Scheme	Marks	AO
Require \(600 - 420(1.0015)^n = 0 \Rightarrow n = \ldots(237.96\ldots)\)	M1	3.1b
So 19 years and 10 months	A1	1.1b
	(2)
	(10 marks)

Notes

M1: Uses a correct strategy to identify the value for \(n\). NB “impossible equations” score M0.

A1: Correct number of years and months.