On the first day of November, Jia deposits $10 000 into a new account which earns 0.4% interest per month, compounded monthly - HSC - SSCE Mathematics Advanced - Question 25 - 2023 - Paper 1

For the expression of $A_n$, we proceed using the geometric series:

Starting from $A_{n-1}$:

$A_{n} = A_{n-1}(1.004) - M$

If we repeatedly apply the withdrawal and interest formula, over $n$ months, we can derive:

$A_n = 10 000(1.004)^n - M(1.004^{n-1} + 1.004^{n-2} + ... + 1)$

The sum of the series can be expressed as:

rac{(1.004^{n} - 1)}{0.004}

Thus, simplifying gives:

A_n = 10 000(1.004)^n - M rac{(1.004^n - 1)}{0.004}

Further manipulation yields:

$A_n = (10 000 - 250M)(1.004)^n + 250M$

To ensure Jia can make at least 100 withdrawals, we establish the condition:

$A_{100} > 0$

Substituting $n = 100$ in $A_n$ gives:

$(10 000 - 250M)(1.004)^{100} + 250M > 0$

Calculating $(1.004)^{100}$ which is approximately $1.48024$:

Bringing terms together:

$10 000(1.48024) - 250M(1.48024) + 250M > 0$

Simplifying leads to:

$14 802.4 - 250M(0.48024) > 0$

Thus,

$250M(0.48024) < 14 802.4$

Solving for $M$ gives:

M < rac{14 802.4}{250 imes 0.48024}

Calculating this results in:

$M < 121.52$

Therefore, the largest amount Jia could withdraw is approximately $121.52$.