Arithmetic and geometric progressions: ASRJC 2024 Promo Q12

Anderson Serangoon Junior College2024 PromoPromo Q12●●● Challenging12 marks

What this question tests

Sequences & Series (APGP).

Question

At a particular bank, a savings account pays an interest of $0.9\%$ per month on the last day of each month. On $1$ January $2024$ Mr Monie put an initial deposit of $\$10000$ into the savings account and continues to make a deposit of $\$810$ at the start of each subsequent month.

Find the total amount in Mr Monie’s savings account at the end of the $2$ months. [1]
Show that the total amount in the savings account at the end of the $n^{\text{th}}$ month is given by $\$100900(1.009)^{n-1} - \$90810$ . [3]
On what date did the value of Mr Monie’s account first exceed $\$30000$ ? [4]

Mr Pey has $\$30000$ and is considering whether to keep this amount (with no further deposit) in the savings account to earn interest or invest this amount in another financial product. This financial product gives a disbursement of $\$100$ at the end of the first month. At the end of every subsequent month, the financial product gives $\$10$ more than the previous month. For example, at the end of the first month, Mr Pey will have $\$30100$ and at the end of two months, he will have $\$30210$ , and so on.

Find, in terms of $N$ , the total amount Mr Pey will have at the end of $N$ months if he invests in the financial product. [2]
Given that $0 < N < 400$ , find the range of values of $N$ such that investing in the financial product is more profitable than keeping the money in the savings account. [2]

Show full worked solution▾

(a)

$10000(1.009)^2+810(1.009)=\$10998.10$

(b)

Month	Start of month	End of month
$1$	$10000$	$10000(1.009)$
$2$	$10000(1.009)+810$	$10000(1.009)^2+810(1.009)$
$3$	$10000(1.009)^2+810(1.009)+810$	$10000(1.009)^3+810(1.009)^2+810(1.009)$
$\vdots$	$\vdots$	$\vdots$
$n$

At the end of nth month, $\begin{aligned} S_n&=10000(1.009)^n+810(1.009)^{n-1}+\dots+810(1.009)\\ &=\$10000(1.009)^n+\$810\left(\frac{(1.009)[(1.009)^{n-1}-1]}{(1.009)-1}\right)\\ &=\$10000(1.009)^n+\$90810[(1.009)^{n-1}-1]\\ &=\$10000(1.009)(1.009)^{n-1}+\$90810(1.009)^{n-1}-\$90810\\ &=\$100900(1.009)^{n-1}-\$90810\qquad(\text{Shown}) \end{aligned}$

(c)

$\begin{aligned} 100900(1.009)^{n-1}-90810&>30000\\ 100900(1.009)^{n-1}&>90810+30000\\ n-1&>\frac{\lg1.197324}{\lg1.009}\\ n&>21.0998 \end{aligned}$

OR using GC

	Amt at the end of the month
$n$	$100900(1.009)^{n-1}-90810$
$21$	$29892.01$
$22$	$30978.32$
$23$	$32074.42$

When $n=21$ , at the end of the $21^{\mathrm{st}}$ month, amount in account $=\$29892.01$

At the start of the $22^{\mathrm{nd}}$ month, amount in account $=\$29892.01+\$810=\$30702.01$

$\therefore$ on $1$ Oct $2025$ , the account will first exceed $\$30000$ .

(d)

Amount Mr. P will have at the end of $N$ months $\begin{aligned} &=30000+[100+110+\dots+(100+10(N-1))]\\ &=30000+\frac N2[2(100)+(N-1)(10)]\\ &=30000+95N+5N^2 \end{aligned}$

(e)

$30000+95N+5N^2>30000\times1.009^N$

$Anderson Serangoon Junior College 2024 Promo Q12: solution diagram$

	$30000+95N+5N^2$	$30000\times1.009^N$
$\vdots$	$\vdots$	$\vdots$
$48$	$46080$	$46120.84$
$49$	$46660$	$46535.93$
$50$	$47250$	$46954.75$
$\vdots$	$\vdots$	$\vdots$
$343$	$650830$	$648322.77$
$344$	$654360$	$654157.67$
$345$	$657900$	$660045.09$

From GC, $49\leq N\leq344$

Answers

(a) $\$10998.10$ (c) $1$ October $2025$
(d) $30000 + 95N + 5N^2$ (e) $49 \leq N \leq 344$

← All Arithmetic and geometric progressions questions JC H2 Maths tuition at Math Academy