Progressions Worked Examples (KBAT)

Q: What is the quickest way to add the 11th to the 20th terms of an AP?

Use S20−S10S_{20}-S_{10}S20−S10. Adding ten terms one by one is slow and error-prone; the difference of two partial sums gives the block total directly.

Q: Why did we reject d=0 in the first example?

A common difference of 000 makes every term equal to 444, which is a geometric progression only in the trivial sense of ratio 111. The intended non-trivial progression comes from d=4d=4d=4.

What these examples cover

These hard Progressions examples ask you to hold two ideas in the same question. One links an arithmetic progression to a geometric progression through the condition that three terms are in geometric ratio; one uses the sum to infinity to find a tail, the sum of every term beyond a certain point; and one finds the total of a block of terms as the difference of two partial sums.

The numbers are still deliberately clean, so the challenge is the reasoning, not the calculator work. Attempt each in full on paper first, then check line by line.

Where your working parts from ours, that step is the real target, hard questions usually turn on a single idea, cleanly applied.

Worked examples

Work through all three. Each rewards a clear plan before any algebra: name what you know, name what you want, and choose the one relationship that connects them.

Attempt each fully before reading the matching solution.

Q1[6 marks]

The first, second and fourth terms of an arithmetic progression, whose first term is $4$ , are also three consecutive terms of a geometric progression. Find (a) the common difference of the AP, (b) the common ratio of the GP, and (c) the sum of the first $5$ terms of that GP.

Show worked solution

(a) With first term $a=4$ and common difference $d$ , the first, second and fourth AP terms are $4$ , $4+d$ and $4+3d$ . Three consecutive GP terms satisfy "middle squared equals the product of the outer two", so:

(4+d)^{2}=4\,(4+3d)

Expand both sides and simplify:

16+8d+d^{2}=16+12d

d^{2}+8d-12d=0 \;\Rightarrow\; d^{2}-4d=0 \;\Rightarrow\; d(d-4)=0

So $d=0$ or $d=4$ . A common difference of $0$ makes every term equal, a trivial ratio of $1$ , so we take $d=4$ .

(b) The three GP terms are now $4$ , $4+4=8$ and $4+3(4)=16$ . The common ratio is any term over the one before:

r=\frac{8}{4}=2 \qquad \left(\text{check: } \frac{16}{8}=2\right)

S_{5}=\frac{4(2^{5}-1)}{2-1}=\frac{4(32-1)}{1}=4(31)=124

Answer

$d=4$ , $r=2$ , and the GP's first five terms sum to $124$ . Check by listing the GP $4,8,16,32,64$ : the total is $4+8+16+32+64=124$ , and the terms $4,8,16$ do sit at positions $1,2,4$ of the AP $4,8,12,16,\ldots$

Q2[5 marks]

A geometric progression has first term $18$ and a sum to infinity of $27$ . Find (a) the common ratio, and (b) the sum of all the terms from the fourth term onwards.

Show worked solution

(a) The sum to infinity is $S_{\infty}=\dfrac{a}{1-r}$ , valid here because the sum exists. Substitute $a=18$ and $S_{\infty}=27$ , then solve for $r$ :

27=\frac{18}{1-r} \;\Rightarrow\; 1-r=\frac{18}{27}=\frac{2}{3} \;\Rightarrow\; r=\frac{1}{3}

Since $|r|=\tfrac{1}{3}<1$ , the sum to infinity was indeed valid.

(b) The terms from the fourth onwards are everything left after the first three, so their total is $S_{\infty}$ minus the sum of the first three terms. First find $S_{3}$ with $S_{n}=\dfrac{a(1-r^{n})}{1-r}$ (the tidy form when $r<1$ ):

S_{3}=\frac{18\left(1-\left(\frac{1}{3}\right)^{3}\right)}{1-\frac{1}{3}}=\frac{18\left(1-\frac{1}{27}\right)}{\frac{2}{3}}=\frac{18\cdot\frac{26}{27}}{\frac{2}{3}}=26

Now subtract from the sum to infinity:

S_{\infty}-S_{3}=27-26=1

Answer

$r=\tfrac{1}{3}$ , and the terms from the fourth onwards sum to $1$ . Check by treating the fourth term as the start of a new GP: its first term is $T_{4}=18\left(\tfrac{1}{3}\right)^{3}=\tfrac{2}{3}$ , still with ratio $\tfrac{1}{3}$ , so its sum to infinity is $\dfrac{2/3}{1-1/3}=\dfrac{2/3}{2/3}=1$ , which agrees.

Q3[6 marks]

An arithmetic progression is such that its 5th term is $17$ and the sum of its first $10$ terms is $185$ . Find (a) the first term and the common difference, and (b) the sum of the 11th to the 20th terms inclusive.

Show worked solution

(a) Turn each fact into an equation. The 5th term uses $T_{n}=a+(n-1)d$ ; the sum uses $S_{n}=\frac{n}{2}[\,2a+(n-1)d\,]$ :

T_{5}=a+4d=17

S_{10}=\frac{10}{2}\,[\,2a+9d\,]=5(2a+9d)=185 \;\Rightarrow\; 2a+9d=37

From the first equation, $a=17-4d$ . Substitute into $2a+9d=37$ :

2(17-4d)+9d=37 \;\Rightarrow\; 34-8d+9d=37 \;\Rightarrow\; d=3

Then $a=17-4(3)=5$ .

(b) The 11th-to-20th terms are the first $20$ terms with the first $10$ removed, so their total is $S_{20}-S_{10}$ . Compute $S_{20}$ :

S_{20}=\frac{20}{2}\,[\,2(5)+19(3)\,]=10\,[\,10+57\,]=10(67)=670

We already know $S_{10}=185$ , so:

S_{20}-S_{10}=670-185=485

Answer

$a=5$ , $d=3$ , and the 11th-to-20th terms sum to $485$ . Check independently: those ten terms run from $T_{11}=5+10(3)=35$ to $T_{20}=5+19(3)=62$ , so their sum is $\frac{10}{2}(35+62)=5(97)=485$ , which agrees.

Q4[5 marks]

Three consecutive terms of an arithmetic progression have a sum of $24$ and a product of $312$ . Find the possible values of the three terms, and state the common difference in each case.

Show worked solution

Because the three terms are consecutive, write them symmetrically as $a-d$ , $a$ , $a+d$ , their sum then simplifies immediately, without $d$ appearing at all.

(a-d)+a+(a+d)=24

3a=24 \;\Rightarrow\; a=8

Now use the product. With $a=8$ , the product of the three terms is $(a-d)(a)(a+d)=a(a^{2}-d^{2})$ :

8\left(8^{2}-d^{2}\right)=312

64-d^{2}=39 \;\Rightarrow\; d^{2}=25 \;\Rightarrow\; d=\pm5

Both signs of $d$ are valid, they simply describe the same three numbers written in increasing or decreasing order.

Answer

The three terms are $3$ , $8$ , $13$ (with $d=5$ ), or the same numbers in reverse, $13$ , $8$ , $3$ (with $d=-5$ ). Check: $3+8+13=24$ and $3\times8\times13=312$ , both correct.

Q5[5 marks]

The number of bacteria in a culture is $5$ at the start of an experiment. The population triples every hour, so the counts taken at the start of each successive hour, $T_{1}=5, T_{2}, T_{3}, \ldots$ , form a geometric progression with common ratio $3$ .

Find the least number of complete hours after the start of the experiment needed for the population to exceed $10\,000$ .

Show worked solution

With $a=5$ and $r=3$ , the term $T_{n}=ar^{n-1}$ gives the count taken $(n-1)$ hours after the start. We need the least $n$ for which $T_{n}$ exceeds $10\,000$ ; because $n$ sits in the exponent, this needs logarithms.

5(3)^{n-1} > 10\,000

3^{n-1} > 2000

Take $\lg$ of both sides. Since the base $3>1$ , the inequality keeps its direction:

(n-1)\lg 3 > \lg 2000 \;\Rightarrow\; n-1 > \frac{\lg 2000}{\lg 3}

n-1 > 6.919\ldots \;\Rightarrow\; n > 7.919\ldots

$n$ must be a whole number, so take the least integer above $7.919\ldots$ :

n=8

$T_{8}$ is the count taken $8-1=7$ hours after the start, so $7$ complete hours are needed.

Answer

$7$ complete hours. Check: $T_{7}=5(3)^{6}=3645$ (still under $10\,000$ ), while $T_{8}=5(3)^{7}=10\,935$ , which is the first count to exceed $10\,000$ .

Q6[6 marks]

Aiman is choosing between two 10-day pocket-money schemes from his uncle. Scheme A pays RM10 on day 1, increasing by RM10 on each following day (an arithmetic progression).

Scheme B pays RM1 on day 1, doubling on each following day (a geometric progression). (a) Find the total paid under each scheme over the $10$ days.

(b) State which scheme pays more in total, and find the difference.

Show worked solution

Scheme A is an AP with $a=10$ , $d=10$ , $n=10$ . Find its total with $S_{n}=\dfrac{n}{2}\,[\,2a+(n-1)d\,]$ :

S_{10}^{A}=\frac{10}{2}\,[\,2(10)+9(10)\,]=5(20+90)=5(110)=550

Scheme B is a GP with $a=1$ , $r=2$ , $n=10$ . Since $r>1$ , use $S_{n}=\dfrac{a(r^{n}-1)}{r-1}$ :

S_{10}^{B}=\frac{1\left(2^{10}-1\right)}{2-1}=2^{10}-1=1024-1=1023

Compare the two totals:

S_{10}^{B}-S_{10}^{A}=1023-550=473

Answer

Scheme A totals RM550 and Scheme B totals RM1023, so Scheme B pays more, by RM473. Sense check: under Scheme B, day $10$ alone pays $2^{9}=$ RM512, already almost the whole of Scheme A's $10$ -day total, so the large gap is reasonable.

Q7[5 marks]

Three numbers are inserted between $4$ and $324$ so that the five numbers form consecutive terms of a geometric progression. (a) Find the possible value(s) of the common ratio.

(b) Taking the positive common ratio, find the three inserted numbers and their sum.

Show worked solution

With first term $T_{1}=4$ and fifth term $T_{5}=324$ , use $T_{5}=ar^{4}$ to form an equation in $r$ :

324=4r^{4}

r^{4}=81 \;\Rightarrow\; r=\pm3

Both roots satisfy the equation because the power is even; either gives a valid GP through $4$ and $324$ . Taking the positive ratio $r=3$ for part (b):

T_{2}=4(3)=12,\quad T_{3}=12(3)=36,\quad T_{4}=36(3)=108

12+36+108=156

Answer

$r=3$ or $r=-3$ ; with the positive ratio, the three inserted numbers are $12$ , $36$ , $108$ , summing to $156$ . Check: $4,12,36,108,324$ is a GP with ratio $3$ , and $4(3)^{4}=4(81)=324$ as required.

Q8[6 marks]

The sum of the first $n$ terms of an arithmetic progression is given by $S_{n}=3n^{2}+5n$ . (a) Find the first term and the common difference.

(b) Express $T_{n}$ in terms of $n$ . (c) Determine whether $100$ is a term of this progression.

Show worked solution

(a)–(b) The first term is simply $S_{1}$ ; every later term is $T_{n}=S_{n}-S_{n-1}$ .

T_{1}=S_{1}=3(1)^{2}+5(1)=8

T_{n}=S_{n}-S_{n-1}=\left(3n^{2}+5n\right)-\left[3(n-1)^{2}+5(n-1)\right]

T_{n}=6n+2

Check this formula at $n=1$ : $6(1)+2=8$ , which matches $T_{1}$ exactly, so it holds for every $n\ge1$ . Since $T_{n}$ is linear in $n$ with coefficient $6$ , the first term is $a=8$ and the common difference is $d=6$ .

6n+2=100 \;\Rightarrow\; n=\frac{98}{6}=\frac{49}{3}

$n$ is not a whole number, so $100$ is not a term of this progression.

Answer

$a=8$ , $d=6$ , $T_{n}=6n+2$ ; $100$ is not a term (it would need $n=\tfrac{49}{3}$ ). Check: $T_{16}=6(16)+2=98$ and $T_{17}=6(17)+2=104$ , so the progression steps straight over $100$ .

Across all three, the hard part was never the arithmetic, it was choosing the one relationship that unlocks the question: the GP condition on three terms, the tail as $S_{\infty}-S_{3}$ , and the block as $S_{20}-S_{10}$ . Name what connects your knowns to your unknown, and the calculation follows.

Key method points

These three examples show how the same handful of formulas combine into harder questions. Keep these ideas ready for problems that mix two topics.

Three numbers are in geometric progression exactly when the middle one squared equals the product of the outer two: $b^{2}=ac$ .
When a condition gives $d(d-4)=0$ , reject the root that makes the progression trivial (here $d=0$ ) and keep the meaningful one.
The sum of all terms beyond the $k$ th of an infinite GP is $S_{\infty}-S_{k}$ ; equivalently, it is a new GP starting at $T_{k+1}$ .
The sum of a block of AP terms, say the 11th to 20th, is $S_{20}-S_{10}$ , no need to add ten terms one by one.
Two facts about an AP give two equations in $a$ and $d$ ; solve them together before answering the final part.
Because marking is analytic, each clear line, the set-up, the elimination, the substitution, can earn its own method mark.

How a teacher helps

Hard questions rarely fail on arithmetic; they fail on the plan, not spotting that three terms in GP give $b^{2}=ac$ , or not seeing a tail as $S_{\infty}-S_{3}$ . In a one-to-one lesson our teacher helps you name the connecting relationship first, so the algebra becomes routine rather than a guess.

We also rehearse rejecting trivial roots and checking by a second route, the habits that protect full marks. Because our teachers are experienced, you learn the reasoning, not just the answer.

Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same either way.

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Frequently asked questions

How do I use the fact that three terms are in geometric progression?

Set the middle term squared equal to the product of the outer two: if the terms are $a$ , $b$ , $c$ then $b^{2}=ac$ . This gives one equation you can solve for the unknown.

How do I find the sum of every term after a certain point in an infinite GP?

Subtract the partial sum from the sum to infinity. The terms from the fourth onwards total $S_{\infty}-S_{3}$ .

You can check by summing the new GP that starts at the fourth term.

What is the quickest way to add the 11th to the 20th terms of an AP?

Use $S_{20}-S_{10}$ . Adding ten terms one by one is slow and error-prone; the difference of two partial sums gives the block total directly.

Why did we reject

d=0

in the first example?

A common difference of $0$ makes every term equal to $4$ , which is a geometric progression only in the trivial sense of ratio $1$ . The intended non-trivial progression comes from $d=4$ .

Source:SRC-DSKP-EN

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

Progressions, Worked Examples (KBAT)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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