Find an expression for ${A_1}$ and show that ${A_2} = {1.004^2}x + 1.004x$.

(i)     Write down a similar expression for ${A_3}$ and ${A_4}$.

(ii)     Hence show that the amount in Mary’s account the day before she turned 10 years old is given by $251({1.004^{120}} - 1)x$.

Write down an expression for ${A_n}$ in terms of $x$ on the day before Mary turned 18 years old showing clearly the value of $n$.

Mary’s grandparents wished for the amount in her account to be at least $\$ 20\,000$ the day before she was 18. Determine the minimum value of the monthly deposit $\$ x$ required to achieve this. Give your answer correct to the nearest dollar.

As soon as Mary was 18 she decided to invest $\$ 15\,000$ of this money in an account of the same type earning 0.4% interest per month. She withdraws $\$ 1000$ every year on her birthday to buy herself a present. Determine how long it will take until there is no money in the account.

${A_1} = 1.004x$    A1

${A_2} = 1.004(1.004x + x)$    A1

$= {1.004^2}x + 1.004x$    AG

Note: Accept an argument in words for example, first deposit has been in for two months and second deposit has been in for one month.

[2 marks]

(i)     ${A_3} = 1.004({1.004^2}x + 1.004x + x) = {1.004^3}x + {1.004^2}x + 1.004x$     (M1)A1

${A_4} = {1.004^4}x + {1.004^3}x + {1.004^2}x + 1.004x$    A1

(ii)     ${A_{120}} = ({1.004^{120}} + {1.004^{119}} +  \ldots  + 1.004)x$     (A1)

$= \frac{{{{1.004}^{120}} - 1}}{{1.004 - 1}} \times 1.004x$    M1A1

$= 251({1.004^{120}} - 1)x$    AG

[6 marks]

${A_{216}} = 251({1.004^{216}} - 1)x{\text{ }}\left( { = x\sum\limits_{t = 1}^{216} {{{1.004}^t}} } \right)$    A1

[1 mark]

$251({1.004^{216}} - 1)x = 20\,000 \Rightarrow x = 58.22 \ldots$    (A1)(M1)(A1)

Note: Award (A1) for $251({1.004^{216}} - 1)x > 20\,000$, (M1) for attempting to solve and (A1) for $x > 58.22 \ldots$.

$x = 59$    A1

Note: Accept $x = 58$. Accept $x \geqslant 59$.

[4 marks]

$r = {1.004^{12}}{\text{ }}( = 1.049 \ldots )$    (M1)

$15\,000{r^n} - 1000\frac{{{r^n} - 1}}{{r - 1}} = 0 \Rightarrow n = 27.8 \ldots$    (A1)(M1)(A1)

Note: Award (A1) for the equation (with their value of $r$), (M1) for attempting to solve for $n$ and (A1) for $n = 27.8 \ldots$

$n = 28$    A1

Note: Accept $n = 27$.

[5 marks]