Find the amount Phil would owe the bank after 20 years. Give your answer to the nearest dollar.

Show that the total value of Phil’s savings after 20 years is $\frac{{({{1.02}^{20}} - 1)P}}{{(1.02 - 1)}}$.

Given that Phil’s aim is to own the house after 20 years, find the value for $P$ to the nearest dollar.

David wishes to withdraw $5000 at the end of each year for a period of $n$ years. Show that an expression for the minimum value of $Q$ is

$\frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} +  \ldots  + \frac{{5000}}{{{{1.028}^n}}}$.

Hence or otherwise, find the minimum value of $Q$ that would permit David to withdraw annual amounts of $5000 indefinitely. Give your answer to the nearest dollar.

$150000 \times {1.035^{20}}$     (M1)(A1)

$= \$ 298468$     A1

Note:     Only accept answers to the nearest dollar. Accept $298469.

[3 marks]

attempt to look for a pattern by considering 1 year, 2 years etc     (M1)

recognising a geometric series with first term $P$ and common ratio 1.02     (M1)

EITHER

$P + 1.02P +  \ldots  + {1.02^{19}}P{\text{ }}\left( { = P(1 + 1.02 +  \ldots  + {{1.02}^{19}})} \right)$     A1

OR

explicitly identify ${u_1} = P,{\text{ }}r = 1.02$ and $n = 20$ (may be seen as ${S_{20}}$).     A1

THEN

${s_{20}} = \frac{{({{1.02}^{20}} - 1)P}}{{(1.02 - 1)}}$     AG

[3 marks]

$24.297 \ldots P = 298468$     (M1)(A1)

$P = 12284$     A1

Note:     Accept answers which round to 12284.

[3 marks]

METHOD 1

$Q({1.028^n}) = 5000(1 + 1.028 + {1.028^2} + {1.028^3} +  \ldots  + {1.028^{n - 1}})$     M1A1

$Q = \frac{{5000\left( {1 + 1.028 + {{1.028}^2} + {{1.028}^3} + ... + {{1.028}^{n - 1}}} \right)}}{{{{1.028}^n}}}$    A1

$= \frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} +  \ldots  + \frac{{5000}}{{{{1.028}^n}}}$     AG

METHOD 2

the initial value of the first withdrawal is $\frac{{5000}}{{1.028}}$     A1

the initial value of the second withdrawal is $\frac{{5000}}{{{{1.028}^2}}}$     R1

the investment required for these two withdrawals is $\frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}}$     R1

$Q = \frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} +  \ldots  + \frac{{5000}}{{{{1.028}^n}}}$     AG

[3 Marks]

sum to infinity is $\frac{{\frac{{5000}}{{1.028}}}}{{1 - \frac{1}{{1.028}}}}$     (M1)(A1)

$= 178571.428 \ldots$

so minimum amount is $178572     A1

Note:     Accept answers which round to $178571 or $178572.

[3 Marks]