Why is the exponent n-1 in T_{n}=ar^{\,n-1}?

Because the first term aaa has been multiplied by rrr zero times, so T1=ar0=aT_{1}=ar^{0}=aT1=ar0=a. The second term has one factor of rrr, the third has two, and the nnnth has n−1n-1n−1.

Progressions Worked Examples (Easy)

Q: How do I tell an arithmetic progression from a geometric one?

Check what stays constant. If each term minus the previous one is the same, it is arithmetic with common difference ddd. If each term divided by the previous one is the same, it is geometric with common ratio rrr.

Q: When can I use the sum to infinity formula?

Only when ∣r∣<1|r|<1∣r∣<1, so the terms shrink towards zero. Then S∞=a1−rS_{\infty}=\frac{a}{1-r}S∞=1−ra. If ∣r∣≥1|r|\ge 1∣r∣≥1 the terms do not shrink and there is no finite sum.

Q: Do I need to memorise these formulas?

The four main progression formulas are printed on the SPM formula list, but knowing them by heart saves time and helps you choose the right one quickly. Practise until writing the correct formula is automatic.

What these examples cover

These easy Progressions examples build the four moves that carry the whole chapter: recognising an arithmetic progression from its constant common difference and finding any term, adding a block of AP terms with the sum formula, recognising a geometric progression from its constant ratio and finding a term, and summing a shrinking GP to infinity. Every question uses small, clean numbers so you can follow each line without leaning on a calculator.

Use the set the honest way: cover the solution, attempt the question fully on paper, and only then check line by line. Where your working differs from ours, find the exact step that parted, that single line is usually where the real learning sits.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and notice how the same routine runs underneath every one: name the first term, name the fixed step ( $d$ or $r$ ), write the formula, then substitute one careful line at a time.

Q1[3 marks]

A sequence begins $3,\ 7,\ 11,\ 15,\ \ldots$ and continues with the same rule. Find (a) the common difference, and (b) the 10th term.

Show worked solution

(a) The sequence rises by the same amount at each step, so it is an arithmetic progression. The common difference $d$ is any term minus the one before it:

d=7-3=4

(b) With first term $a=3$ and $d=4$ , the $n$ th term is $T_{n}=a+(n-1)d$ . Put $n=10$ :

T_{10}=3+(10-1)(4)=3+9(4)=3+36=39

Answer

The common difference is $4$ and the 10th term is $39$ . Check by counting on: $3,7,11,15,19,23,27,31,35,39$ , the tenth number is indeed $39$ .

Q2[3 marks]

An arithmetic progression has first term $5$ and common difference $3$ . Find the sum of the first $12$ terms.

Show worked solution

The sum of the first $n$ terms of an AP is $S_{n}=\frac{n}{2}\,[\,2a+(n-1)d\,]$ . Here $a=5$ , $d=3$ and $n=12$ .

Substitute, then simplify inside the bracket before you multiply:

S_{12}=\frac{12}{2}\,[\,2(5)+(12-1)(3)\,]

S_{12}=6\,[\,10+33\,]=6(43)=258

Answer

The sum of the first $12$ terms is $258$ . Check with the other form $S_{n}=\frac{n}{2}(a+l)$ : the last term is $l=T_{12}=5+11(3)=38$ , so $S_{12}=6(5+38)=6(43)=258$ , which agrees.

Q3[3 marks]

A sequence begins $2,\ 6,\ 18,\ 54,\ \ldots$ . Find (a) the common ratio, and (b) the 6th term.

Show worked solution

(a) Each term is a fixed multiple of the one before, so this is a geometric progression. The common ratio $r$ is any term divided by the one before it:

r=\frac{6}{2}=3

(b) With first term $a=2$ and $r=3$ , the $n$ th term is $T_{n}=ar^{\,n-1}$ . Put $n=6$ , so the exponent is $6-1=5$ :

T_{6}=2\times 3^{\,6-1}=2\times 3^{5}=2\times 243=486

Answer

The common ratio is $3$ and the 6th term is $486$ . Check by listing: $2,6,18,54,162,486$ , the sixth term is $486$ .

Q4[3 marks]

A geometric progression has first term $12$ and common ratio $\tfrac{1}{2}$ . Find its sum to infinity.

Show worked solution

Because $|r|=\tfrac{1}{2}<1$ , the terms shrink towards zero and the running total settles on a finite value. The sum to infinity is $S_{\infty}=\dfrac{a}{1-r}$ .

Substitute $a=12$ and $r=\tfrac{1}{2}$ :

S_{\infty}=\frac{12}{1-\frac{1}{2}}=\frac{12}{\frac{1}{2}}=24

Answer

The sum to infinity is $24$ . This is believable from the running total $12+6+3+1.5+\ldots$ , which creeps up towards $24$ without ever passing it.

Q5[3 marks]

The numbers $3x-1$ , $x+7$ , and $5x-3$ are three consecutive terms of an arithmetic progression. Find the value of $x$ .

Show worked solution

In an arithmetic progression the gap between consecutive terms is constant, so the middle term sits exactly halfway between its neighbours: twice the middle term equals the sum of the two terms on either side.

2(x+7)=(3x-1)+(5x-3)

Expand both sides and collect the $x$ terms on one side:

2x+14=8x-4 \Rightarrow 18=6x \Rightarrow x=3

Answer

$x=3$ . Check: the terms become $8,10,12$ , a genuine AP with common difference $2$ .

Q6[3 marks]

The numbers $3$ , $x$ , and $27$ are three consecutive terms of a geometric progression. Find the possible value(s) of $x$ .

Show worked solution

In a geometric progression the ratio between consecutive terms is constant, so the middle term squared equals the product of the two terms on either side.

x^{2}=(3)(27)=81

Take the square root of both sides, remember a common ratio can be negative, so both roots are valid here:

x=\pm 9

Answer

$x=9$ or $x=-9$ . With $x=9$ the terms $3,9,27$ form a GP with $r=3$ ; with $x=-9$ the terms $3,-9,27$ form a GP with $r=-3$ , both check out.

Q7[3 marks]

An arithmetic progression has first term $4$ and common difference $5$ . Which term of the progression is equal to $79$ ?

Show worked solution

Use $T_{n}=a+(n-1)d$ with $a=4$ , $d=5$ , and set $T_{n}=79$ , then solve for $n$ :

79=4+(n-1)(5)

Simplify and isolate $n$ :

75=5(n-1) \Rightarrow 15=n-1 \Rightarrow n=16

Answer

$79$ is the 16th term. Check: $T_{16}=4+(15)(5)=4+75=79$ .

Q8[3 marks]

A geometric progression has first term $5$ and common ratio $2$ . Which term of the progression is equal to $160$ ?

Show worked solution

Use $T_{n}=ar^{\,n-1}$ with $a=5$ , $r=2$ , and set $T_{n}=160$ , then solve for $n$ :

160=5\times 2^{\,n-1}

Divide both sides by $5$ , then write $32$ as a power of $2$ so the exponents can be matched:

32=2^{\,n-1}=2^{5} \Rightarrow n-1=5 \Rightarrow n=6

Answer

$160$ is the 6th term. Check: $T_{6}=5\times 2^{5}=5\times 32=160$ .

Look at how different these four appear, yet how alike the discipline is: decide AP or GP first, write down $a$ with $d$ or $r$ , pick the matching formula, then substitute in clean, separate lines. That order is what turns Progressions into a dependable block of early marks.

Key method points

These four examples rehearse the everyday skills that open almost every Progressions question in Add Math. Keep the following points in mind as you practise more.

Test for an AP by checking that consecutive differences are equal; the common difference is $d=T_{2}-T_{1}$ .
Test for a GP by checking that consecutive ratios are equal; the common ratio is $r=\frac{T_{2}}{T_{1}}$ .
Use $T_{n}=a+(n-1)d$ for an AP term and $T_{n}=ar^{\,n-1}$ for a GP term, the exponent is $n-1$ , not $n$ .
Add AP terms with $S_{n}=\frac{n}{2}[\,2a+(n-1)d\,]$ , or $S_{n}=\frac{n}{2}(a+l)$ when the last term $l$ is known.
The sum to infinity $S_{\infty}=\frac{a}{1-r}$ applies only when $|r|<1$ ; otherwise no finite sum exists.
Because marking is analytic, a correct formula line with a clear substitution can earn method marks even if the final arithmetic slips.

How a teacher helps

When a student drops a mark on questions like these, it is usually a small, fixable habit, writing $r^{n}$ instead of $r^{\,n-1}$ , or reaching for the sum-to-infinity formula when $|r|\ge 1$ . In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it before it hardens into a routine.

Because our teachers are experienced, you work with someone who explains why each formula fits. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

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Frequently asked questions

How do I tell an arithmetic progression from a geometric one?

Check what stays constant. If each term minus the previous one is the same, it is arithmetic with common difference $d$ .

If each term divided by the previous one is the same, it is geometric with common ratio $r$ .

Why is the exponent

n-1

T_{n}=ar^{\,n-1}

Because the first term $a$ has been multiplied by $r$ zero times, so $T_{1}=ar^{0}=a$ . The second term has one factor of $r$ , the third has two, and the $n$ th has $n-1$ .

When can I use the sum to infinity formula?

Only when $|r|<1$ , so the terms shrink towards zero. Then $S_{\infty}=\frac{a}{1-r}$ .

If $|r|\ge 1$ the terms do not shrink and there is no finite sum.

Do I need to memorise these formulas?

The four main progression formulas are printed on the SPM formula list, but knowing them by heart saves time and helps you choose the right one quickly. Practise until writing the correct formula is automatic.

Source:SRC-DSKP-EN

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

Progressions, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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