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Arithmetic and geometric progressions: YIJC 2024 Promo Q11

Yishun Innova Junior College2024 PromoPromo Q11●●● Challenging12 marks

What this question tests

Arithmetic and geometric progressions.

Question

(a) Zack saves $x\$x on 1 January 2024. On the first day of each subsequent month, he saves $10 more than in the previous month, so that he saves $(x+10)\$(x+10) on 1 February 2024, $(x+20)\$(x+20) on 1 March 2024, and so on. Find the value of xx if he saves $25 000 in total on 1 December 2026. [3]

(b) Dan opens a savings account. He deposits $200 on 1 January 2024 and continues to deposit $200 on the first day of each subsequent month. The interest rate of the savings account is 0.2% per month, so that on the last day of each month the amount in the account on that day is increased by 0.2%.

(i) Find how much the $200 deposited on 1 January 2024 is worth at the end of 31 December 2024. [1]

(ii) Show that, at the end of the nnth month (where January 2024 is the first month, February 2024 is the second month, and so on), the total amount in the account is $100200(1.002n−1)\$100200(1.002^n-1). [4]

(iii) In which month and year will the amount in Dan’s account first exceed $4500? Explain whether this occurs on the first or last day of the month. [4]

Show full worked solution▾

(a)From 1 Jan 2024 to 1 Dec 2026, there are 36 months. 362(2x+(36−1)(10))=25000\frac{36}{2}\bigl(2x+(36-1)(10)\bigr)=25000 2x+350=1388.888892x+350=1388.88889 x=519.44(2 d.p.)x=519.44\quad(2\text{ d.p.})

11(b)(i)At the end of 31 Dec 2024, the first $200 will be worth $200(1.002)12≈$204.85(2 d.p.)\$200(1.002)^{12}\approx\$204.85\quad(2\text{ d.p.})

(ii)Method 1:

MthStartEnd
1200200200(1.002)200(1.002)
2200(1.002)+200200(1.002)+200200(1.0022)+200(1.002)200(1.002^2)+200(1.002)
3200(1.0022)+200(1.002)200(1.002^2)+200(1.002)
+200+200		200(1.0023)+200(1.0022)200(1.002^3)+200(1.002^2)
+200(1.002)+200(1.002)
⋮\vdots⋮\vdots⋮\vdots
nn200(1.002n)+200(1.002n−1)200(1.002^n)+200(1.002^{n-1})
+⋯+200(1.002)+\cdots+200(1.002)

Total amount in the account at end of last day of nthn^{\text{th}} month =200(1.002n)+200(1.002n−1)+⋯+200(1.002)=200(1.002^n)+200(1.002^{n-1})+\cdots+200(1.002) =200[1.002+1.0022+⋯+1.002n]=200\left[1.002+1.002^2+\cdots+1.002^n\right]

GP, a=1.002a=1.002, r=1.002r=1.002, nn terms

=200[1.002(1.002n−1)1.002−1]=200\left[\frac{1.002(1.002^n-1)}{1.002-1}\right] =100200(1.002n−1)(Shown)=100200(1.002^n-1)\quad\text{(Shown)}

Method 2:

At the end of the nnth month,

The 1st1^{\text{st}} $200 deposited is worth $200(1.002)n(1.002)^n

The 2nd2^{\text{nd}} $200 deposited is worth $200(1.002)n−1(1.002)^{n-1}

The 3rd3^{\text{rd}} $200 deposited is worth $200(1.002)n−2(1.002)^{n-2}

⋮\vdots

The nthn^{\text{th}} $200 deposited is worth $200(1.002)(1.002)

Total amount in the account =200[1.002+1.0022+⋯+1.002n]=200\left[1.002+1.002^2+\cdots+1.002^n\right]

GP, a=1.002a=1.002, r=1.002r=1.002, nn terms

=200[1.002(1.002n−1)1.002−1]=200\left[\frac{1.002(1.002^n-1)}{1.002-1}\right] =100200(1.002n−1)(Shown)=100200(1.002^n-1)\quad\text{(Shown)}

(iii) total amount at end of nth month=100200(1.002n−1)>4500\text{total amount at end of $n$th month}=100200(1.002^n-1)>4500 Using GC,

When n=21n=21, total amount =4293.64=4293.64

When n=22n=22, total amount =4502.63=4502.63

Dan’s account will first exceed $4500 in Oct 2025.

Amount at start of Oct 2025 =$4293.64+$200=\$4293.64+\$200 =$4493.64<$4500=\$4493.64<\$4500 It occurs at the end of the month.

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