Progressions Worked Examples (Medium)

Q: Given two terms of an AP, how do I find a and d?

Write each term with Tn=a+(n−1)dT_{n}=a+(n-1)dTn=a+(n−1)d to get two equations, then subtract one from the other to eliminate aaa and solve for ddd. Substitute ddd back to find aaa.

Q: Why do I divide the two equations for a GP but subtract for an AP?

An AP term is a+(n−1)da+(n-1)da+(n−1)d, so subtracting cancels aaa and leaves a multiple of ddd. A GP term is ar n−1ar^{\,n-1}arn−1, so dividing cancels aaa and leaves a power of rrr.

Q: When an AP sum gives a quadratic in n, which root do I keep?

Keep the positive whole-number root. The number of terms cannot be negative or a fraction, so a solution like n=−212n=-\tfrac{21}{2}n=−221 is rejected on those grounds.

Q: Which version of the GP sum formula should I use?

They are equal, so either works. Use Sn=a(rn−1)r−1S_{n}=\frac{a(r^{n}-1)}{r-1}Sn=r−1a(rn−1) when r>1r>1r>1 and Sn=a(1−rn)1−rS_{n}=\frac{a(1-r^{n})}{1-r}Sn=1−ra(1−rn) when r<1r<1r<1; each keeps the signs positive and tidy.

What these examples cover

These medium Progressions examples are the natural next step once the four core formulas feel automatic. Instead of handing you $a$ , $d$ or $r$ directly, each question hides them one layer down: you are given two terms and must set up a pair of equations, or given a sum and must solve for the number of terms.

The numbers stay small and clean so the algebra, not the arithmetic, is the point. Use the set the honest way, cover the solution, attempt the whole question on paper, and only then check line by line.

Where your route differs from ours, find the exact step that parted; that is where a medium question is won or lost.

Worked examples

Work through all three. Each one rewards the same discipline: turn the given facts into equations, solve for the unknowns cleanly, then answer exactly what was asked.

Attempt each fully before reading the matching solution.

Q1[4 marks]

In an arithmetic progression, the 4th term is $14$ and the 9th term is $34$ . Find (a) the first term and the common difference, and (b) the 20th term.

Show worked solution

(a) Write each given term with $T_{n}=a+(n-1)d$ . The 4th term gives one equation and the 9th term gives another:

T_{4}=a+3d=14 \qquad T_{9}=a+8d=34

Subtract the first equation from the second to remove $a$ :

(a+8d)-(a+3d)=34-14 \;\Rightarrow\; 5d=20 \;\Rightarrow\; d=4

Substitute $d=4$ back into $a+3d=14$ :

a+3(4)=14 \;\Rightarrow\; a=14-12=2

(b) Now use $T_{n}=a+(n-1)d$ with $n=20$ :

T_{20}=2+(20-1)(4)=2+19(4)=2+76=78

Answer

$a=2$ , $d=4$ , and the 20th term is $78$ . Check the given data: $T_{4}=2+3(4)=14$ and $T_{9}=2+8(4)=34$ , both correct.

Q2[4 marks]

An arithmetic progression has first term $3$ and common difference $4$ . The sum of the first $n$ terms is $210$ .

Find the value of $n$ .

Show worked solution

Use the AP sum $S_{n}=\frac{n}{2}\,[\,2a+(n-1)d\,]$ with $a=3$ and $d=4$ , and set it equal to $210$ :

\frac{n}{2}\,[\,2(3)+(n-1)(4)\,]=210

Simplify inside the bracket first: $2(3)+(n-1)(4)=6+4n-4=4n+2$ . So

\frac{n}{2}(4n+2)=210 \;\Rightarrow\; n(2n+1)=210

Expand and bring everything to one side to form a quadratic in $n$ :

2n^{2}+n-210=0

Factorise. Since $2\times(-210)=-420$ and $21\times(-20)=-420$ with $21-20=1$ , the factors are:

(2n+21)(n-10)=0 \;\Rightarrow\; n=-\tfrac{21}{2} \ \text{or}\ n=10

Answer

A number of terms must be a positive whole number, so $n=10$ and we reject $n=-\tfrac{21}{2}$ . Check: $S_{10}=\frac{10}{2}[\,2(3)+9(4)\,]=5(6+36)=5(42)=210$ , as required.

Q3[4 marks]

In a geometric progression, the 2nd term is $6$ and the 5th term is $48$ . Find (a) the common ratio and the first term, and (b) the sum of the first $6$ terms.

Show worked solution

(a) Write each given term with $T_{n}=ar^{\,n-1}$ :

T_{2}=ar=6 \qquad T_{5}=ar^{4}=48

Divide the second equation by the first so $a$ cancels and only $r$ is left:

\frac{ar^{4}}{ar}=\frac{48}{6} \;\Rightarrow\; r^{3}=8 \;\Rightarrow\; r=2

Substitute $r=2$ into $ar=6$ :

a(2)=6 \;\Rightarrow\; a=3

(b) With $a=3$ , $r=2$ and $n=6$ , use $S_{n}=\dfrac{a(r^{n}-1)}{r-1}$ , the neat form when $r>1$ :

S_{6}=\frac{3(2^{6}-1)}{2-1}=\frac{3(64-1)}{1}=3(63)=189

Answer

$r=2$ , $a=3$ , and the sum of the first $6$ terms is $189$ . Check by listing the terms $3,6,12,24,48,96$ : their total is $189$ , and indeed $T_{2}=6$ and $T_{5}=48$ .

Q4[4 marks]

The numbers $3x$ , $x+8$ , and $21$ are three consecutive terms of an arithmetic progression. Find (a) the value of $x$ , and (b) the common difference and the 5th term of the progression.

Show worked solution

For three consecutive AP terms, twice the middle term equals the sum of the outer terms: $2T_{2}=T_{1}+T_{3}$ .

2(x+8)=3x+21

Expand and solve for $x$ :

2x+16=3x+21 \;\Rightarrow\; 16-21=3x-2x \;\Rightarrow\; x=-5

(b) Find $a=3x$ and the common difference $d$ , then use $T_{n}=a+(n-1)d$ for the 5th term:

a=3(-5)=-15 \qquad d=(x+8)-3x=-2(-5)+8=18

T_{5}=a+4d=-15+4(18)=-15+72=57

Answer

$x=-5$ . The three terms are $-15$ , $3$ , $21$ (common difference $18$ ), and $T_{5}=57$ .

Check: $2(3)=6=-15+21$ .

Q5[4 marks]

A geometric progression has second term $6$ and sum to infinity $27$ . Find the possible values of the first term $a$ and the common ratio $r$ .

Show worked solution

Use $T_{2}=ar=6$ and $S_{\infty}=\dfrac{a}{1-r}=27$ (valid since the series converges, so $|r|<1$ ). From the $S_{\infty}$ equation, $a=27(1-r)$ .

a=27(1-r)

Substitute into $ar=6$ :

27r(1-r)=6 \;\Rightarrow\; 27r-27r^{2}=6 \;\Rightarrow\; 9r^{2}-9r+2=0

Factorise:

(3r-2)(3r-1)=0 \;\Rightarrow\; r=\tfrac{2}{3} \ \text{or}\ r=\tfrac{1}{3}

Substitute each $r$ back into $a=27(1-r)$ . Both satisfy $|r|<1$ , so both are valid:

r=\tfrac{2}{3}: a=27(1-\tfrac{2}{3})=9 \qquad r=\tfrac{1}{3}: a=27(1-\tfrac{1}{3})=18

Answer

Two progressions fit: $a=9$ , $r=\tfrac{2}{3}$ or $a=18$ , $r=\tfrac{1}{3}$ . Check: $9\times\tfrac{2}{3}=6$ and $18\times\tfrac{1}{3}=6$ , both matching $T_{2}=6$ ; and $\tfrac{9}{1-2/3}=27$ , $\tfrac{18}{1-1/3}=27$ , both matching $S_{\infty}=27$ .

Q6[4 marks]

The sum of the first $n$ terms of an arithmetic progression is given by $S_{n}=3n^{2}+5n$ . Find (a) the first term, and (b) the 6th term of the progression.

Show worked solution

(a) The first term equals $S_{1}$ (the sum of just one term):

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

(b) For $n\ge2$ , the $n$ th term is $S_{n}-S_{n-1}$ . Use this to find $T_{6}$ as $S_{6}-S_{5}$ :

S_{6}=3(6)^{2}+5(6)=108+30=138 \qquad S_{5}=3(5)^{2}+5(5)=75+25=100

T_{6}=S_{6}-S_{5}=138-100=38

Answer

The first term is $8$ , and $T_{6}=38$ . Check with the general term $T_{n}=S_{n}-S_{n-1}=6n+2$ (found by expanding): $T_{1}=6(1)+2=8$ and $T_{6}=6(6)+2=38$ , both matching.

Q7[4 marks]

In a geometric progression, the first term is $3$ and the common ratio is $2$ . Find the smallest value of $n$ for which $T_{n}$ exceeds $300$ .

Show worked solution

Write $T_{n}$ using the GP formula and set up the inequality:

T_{n}=3(2)^{n-1} > 300 \;\Rightarrow\; 2^{n-1} > 100

Take $\lg$ of both sides ( $n-1$ is now an exponent, so $\lg$ brings it down):

(n-1)\lg 2 > \lg 100 \;\Rightarrow\; n-1 > \frac{2}{\lg 2} \approx \frac{2}{0.3010} \approx 6.64

Since $n$ must be a whole number, round up to the next integer above $6.64$ : $n-1=7$ , so $n=8$ . Verify by checking the boundary terms:

T_{7}=3(2)^{6}=192\;(\text{not} > 300) \qquad T_{8}=3(2)^{7}=384\;(>300)

Answer

The smallest value is $n=8$ . $T_{7}=192$ does not exceed $300$ , but $T_{8}=384$ does, confirming $n=8$ is the first term past $300$ .

Q8[4 marks]

The numbers $2$ , $x$ , and $x+4$ are three consecutive terms of a geometric progression. Find (a) the possible values of $x$ , and (b) the corresponding values of the common ratio $r$ .

Show worked solution

For three consecutive GP terms, the square of the middle term equals the product of the outer terms:

x^{2}=2(x+4)

Expand and rearrange into a quadratic equation, then factorise:

x^{2}-2x-8=0 \;\Rightarrow\; (x-4)(x+2)=0 \;\Rightarrow\; x=4 \ \text{or}\ x=-2

(b) Find $r=\dfrac{x}{2}$ for each case:

x=4: r=\frac{4}{2}=2 \qquad x=-2: r=\frac{-2}{2}=-1

Answer

$x=4$ gives the GP $2,4,8$ with $r=2$ ; $x=-2$ gives the GP $2,-2,2$ with $r=-1$ . Both are valid geometric progressions (check: $4^{2}=2\times8=16$ , and $(-2)^{2}=2\times2=4$ ).

Notice the shared shape of all three: read the given facts into equations, eliminate one unknown (by subtracting for an AP, by dividing for a GP), solve, then substitute back into the formula the question actually asks for. That habit is what makes medium Progressions questions feel routine.

Key method points

These three examples rehearse the reasoning that most medium Progressions questions rely on. Carry these points into your own practice.

Two terms of an AP give two linear equations in $a$ and $d$ ; subtract to eliminate $a$ and find $d$ first.
Two terms of a GP give two equations in $a$ and $r$ ; divide to eliminate $a$ and find $r$ first.
Setting an AP sum equal to a target usually produces a quadratic in $n$ , form $=0$ , then factorise or use the formula.
A count of terms $n$ must be a positive whole number, so reject negative or fractional solutions.
Use $S_{n}=\frac{a(r^{n}-1)}{r-1}$ when $r>1$ and $S_{n}=\frac{a(1-r^{n})}{1-r}$ when $r<1$ to keep signs tidy.
Always check your answer against the original given terms, it catches a sign slip in seconds.

How a teacher helps

Medium questions are where students either build confidence or quietly lose it, and the difference is often one habit, knowing to subtract for an AP but divide for a GP, or remembering that $n$ cannot be negative. In a one-to-one lesson our teacher watches how you set up the equations and steps in at the exact moment a method wanders, before it becomes a pattern.

Because our teachers are experienced, you work with someone who shows why each move follows. 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

Given two terms of an AP, how do I find

a

and

d

Write each term with $T_{n}=a+(n-1)d$ to get two equations, then subtract one from the other to eliminate $a$ and solve for $d$ . Substitute $d$ back to find $a$ .

Why do I divide the two equations for a GP but subtract for an AP?

An AP term is $a+(n-1)d$ , so subtracting cancels $a$ and leaves a multiple of $d$ . A GP term is $ar^{\,n-1}$ , so dividing cancels $a$ and leaves a power of $r$ .

When an AP sum gives a quadratic in

n

, which root do I keep?

Keep the positive whole-number root. The number of terms cannot be negative or a fraction, so a solution like $n=-\tfrac{21}{2}$ is rejected on those grounds.

Which version of the GP sum formula should I use?

They are equal, so either works. Use $S_{n}=\frac{a(r^{n}-1)}{r-1}$ when $r>1$ and $S_{n}=\frac{a(1-r^{n})}{1-r}$ when $r<1$ ; each keeps the signs positive and tidy.

Source:SRC-DSKP-EN

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

Progressions, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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