KBAT: Progressions in Financial Problems

Q: How do I tell whether a savings story is arithmetic or geometric?

Look at how one year becomes the next. If a fixed amount is added,'RM2000 more each year', it is arithmetic with common difference ddd. If a fixed percentage is applied,'5% more each year', meaning ×1.05\times 1.05×1.05, it is geometric with common ratio rrr. Deciding this first tells you which formulas to use and prevents mixing them up.

Q: Which sum formula do I use for a total, and can I still score if I choose wrong?

Use Sn=n2[2a+(n−1)d]S_n=\tfrac{n}{2}[2a+(n-1)d]Sn=2n[2a+(n−1)d] for an arithmetic total and Sn=a(rn−1)r−1S_n=\tfrac{a(r^{n}-1)}{r-1}Sn=r−1a(rn−1) for a geometric total. Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, so method marks are given line by line, correct substitution and honest working can still earn marks even if you first pick the wrong progression, though naming the type early protects them best.

What makes this a KBAT question

A routine progressions question tells you it is arithmetic or geometric, hands you the first term with a common difference or ratio, and asks for a term or a sum. A financial KBAT question hides all of that inside a money story, two savings plans, a salary with yearly rises, a loan cleared in shrinking instalments, and leaves you to decide which progression each scheme is.

You must translate 'RM2000 more each year' into a common difference and '5% more each year' into a common ratio, then compare and advise. That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: the formulas are ordinary Form 4 work, but modelling the situation, choosing between arithmetic and geometric, and interpreting which plan is really better are not spelt out for you.

The reasoning, not the substitution, is what earns the marks.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[8 marks]

A worker is offered two salary schemes, each running for 10 years. Under Scheme A the annual salary starts at RM30 000 in year 1 and rises by a fixed RM2000 every year.

Under Scheme B the annual salary starts at RM30 000 in year 1 and rises by 5% of the previous year's salary every year. (a) Write an expression for the salary in year $n$ under each scheme.

(b) Find the total amount earned over the 10 years under each scheme, giving Scheme B to the nearest ringgit. (c) State which scheme pays more over the 10 years, then find the first year in which Scheme B's annual salary exceeds Scheme A's, and explain why the scheme with the higher percentage growth does not win over the 10-year period.

Show worked solution

Understand. Scheme A adds a fixed amount each year, so it is an arithmetic progression.

Scheme B multiplies by a fixed factor each year, so it is a geometric progression. Both start at RM30 000.

We need each year's salary, each 10-year total, and a comparison.

Plan. For Scheme A use the arithmetic term and sum formulas with first term $a=30000$ and common difference $d=2000$ .

For Scheme B use the geometric term and sum formulas with $a=30000$ and common ratio $r=1.05$ . Then compare the two totals and test the annual salaries year by year.

Execute. (a) Scheme A, year $n$ :

T_n = a+(n-1)d = 30000+(n-1)(2000) = 28000+2000n

Scheme B, year $n$ :

T_n = ar^{\,n-1} = 30000(1.05)^{\,n-1}

(b) Total for Scheme A over 10 years:

S_{10} = \frac{n}{2}\left[2a+(n-1)d\right] = \frac{10}{2}\left[2(30000)+9(2000)\right] = 5(78000) = 390000

Total for Scheme B over 10 years:

S_{10} = \frac{a(r^{n}-1)}{r-1} = \frac{30000(1.05^{10}-1)}{1.05-1} = 600000\,(1.6288946-1)

S_{10} = 600000(0.6288946) = 377336.78

So Scheme A pays RM390 000 and Scheme B pays about RM377 337 over the 10 years.

(c) Scheme A pays more over the 10 years, by $390000-377337 = \text{RM}12\,663$ . To find when Scheme B's yearly salary first passes Scheme A's, compare year by year.

In year 10 Scheme A pays $28000+2000(10)=\text{RM}48\,000$ while Scheme B pays $30000(1.05)^{9}=\text{RM}46\,540$ , so A is still ahead. Continuing: in year 13, A pays RM54 000 and B pays $30000(1.05)^{12}\approx\text{RM}53\,876$ ; in year 14, A pays RM56 000 and B pays $30000(1.05)^{13}\approx\text{RM}56\,569$ .

So Scheme B's annual salary first exceeds Scheme A's in year 14, beyond the 10-year window.

The percentage scheme loses over 10 years because a fixed RM2000 rise is 6.7% of the starting salary, larger than B's 5%, so early on A climbs faster. B's advantage is compounding, which only overtakes A's steady RM2000 in the long run, here not until year 14.

Over a short, fixed period the larger absolute increment wins.

Check. Year 1 agrees for both: $28000+2000=30000$ and $30000(1.05)^{0}=30000$ .

The two 10-year totals are close (within about RM13 000), which is sensible for schemes that start equal and grow at similar early rates.

Finding a sensible first step

When money is dressed up in words, the reliable first move is to decide, for each scheme, whether the change from one year to the next is a fixed amount added or a fixed percentage multiplied. 'RM2000 more each year' is a constant amount, so the scheme is arithmetic with common difference $d=2000$ .

'5% more each year' means each year is $1.05$ times the last, so the scheme is geometric with common ratio $r=1.05$ . Write the first term $a$ for each and label the type clearly.

Only then reach for a formula: the $n$ th-term formula when the question wants a single year's salary, the sum formula when it wants a total over several years. Naming the progression first prevents the commonest error, using an arithmetic formula on a geometric scheme, or the reverse.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a financial progressions question a marker looks for:

Each scheme correctly identified as arithmetic or geometric, with $a$ and $d$ or $a$ and $r$ stated.
The correct $n$ th-term formula for a single year, $T_n=a+(n-1)d$ or $T_n=ar^{\,n-1}$ .
The correct sum formula for a total, $S_n=\tfrac{n}{2}[2a+(n-1)d]$ or $S_n=\tfrac{a(r^{n}-1)}{r-1}$ .
Accurate substitution and calculator use, especially in evaluating $1.05^{10}$ .
A clear numerical comparison of the two 10-year totals.
A reasoned recommendation and an interpretation, why higher percentage growth need not win over a fixed period.

How a teacher helps

Financial progression questions reward students who model carefully and then explain their choice, and both habits sharpen with feedback. In a one-to-one lesson our teachers ask you to say, in words, whether each scheme adds or multiplies before you touch a formula, to write $a$ , $d$ or $r$ on the page, and to check your calculator working for $r^{n}$ step by step.

We spend time on the interpretation, why a scheme with a higher percentage rise can still lose over ten years, because that sentence earns the reasoning mark. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

How do I tell whether a savings story is arithmetic or geometric?

Look at how one year becomes the next. If a fixed amount is added,'RM2000 more each year', it is arithmetic with common difference $d$ .

If a fixed percentage is applied,'5% more each year', meaning $\times 1.05$ , it is geometric with common ratio $r$ . Deciding this first tells you which formulas to use and prevents mixing them up.

Why does the scheme with faster percentage growth not win over ten years?

Because a fixed RM2000 rise is about 6.7% of the RM30 000 starting salary, larger than the other scheme's 5%, so the arithmetic plan climbs faster at first. The geometric plan grows by compounding, which only overtakes the steady RM2000 much later; in this problem not until year 14.

Over a short, fixed window the larger absolute increment can beat the higher percentage.

Which sum formula do I use for a total, and can I still score if I choose wrong?

Use $S_n=\tfrac{n}{2}[2a+(n-1)d]$ for an arithmetic total and $S_n=\tfrac{a(r^{n}-1)}{r-1}$ for a geometric total. Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, so method marks are given line by line, correct substitution and honest working can still earn marks even if you first pick the wrong progression, though naming the type early protects them best.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026