MathsCBSE Class 9 · Sequences and Progressions

Geometric Progressions and Recursive Sequences: Class 9

Updated 17 October 2026 · Fact-checked

In a geometric progression (GP) each term is the previous term multiplied by a constant common ratio $r$, so $a_n = ar^{n-1}$. A recursive rule defines each term from earlier ones: the Virahānka–Fibonacci sequence uses $u_n = u_{n-1} + u_{n-2}$, and the Tower of Hanoi needs $M_n = 2M_{n-1} + 1$ moves, which equals $2^n - 1$. Fractals such as the Sierpiński triangle and carpet give GPs for the number and size of pieces.

Each stage has 3 times as many pieces as the one before, so the counts 1, 3, 9 form a GP with common ratio 3.

Illustration drawn for QuestionBuddy.

Understand Geometric Progressions, Recursive Rules and Fractal Patterns

Geometric progression. In a GP the ratio of consecutive terms is constant: $r = \frac{a _{n + 1}}{a _{n}}$ . Starting from $a$ , the terms are $a, a r, a r^{2}, \dots$ and the $n$ th term is $a_{n} = a r^{n - 1}$ . For $3, 6, 12, \dots$ : $r = 2$ , $a_{8} = 3 \times 2^{7} = 384$ , and since $3 \times 2^{n - 1} = 768$ gives $2^{n - 1} = 256$ , the term $768$ is the 9th. A GP with $∣ r ∣ > 1$ grows rapidly (doubling every hour), and with $∣ r ∣ < 1$ shrinks (each stage keeping $\frac{3}{4}$ of the area).

From two terms. If $a_{3} = 12$ and $a_{6} = 96$ , then $\frac{a _{6}}{a _{3}} = r^{3} = 8$ , so $r = 2$ and $a = \frac{12}{4} = 3$ . For three numbers in GP with a given sum and product, write them as $\frac{a}{r}$ , $a$ , $a r$ : the product is $a^{3}$ . If the product is $216$ , $a = 6$ ; a sum of $21$ gives $6 (\frac{1}{r} + 1 + r) = 21$ , so $r + \frac{1}{r} = \frac{5}{2}$ , $r = 2$ or $\frac{1}{2}$ , and the numbers are $3, 6, 12$ (or the same in reverse).

AP or GP? To test a sequence, compute differences and ratios. $5, 9, 13, 17$ has constant differences ( $d = 4$ ): an AP. $2, 6, 18, 54$ has constant ratios ( $r = 3$ ): a GP. $1, 4, 9, 16$ has differences $3, 5, 7$ and ratios $4, 2.25, \dots$ : neither. Doubling of a population or a message forwarded to 3 new people each round are GPs: after $n$ rounds, $3^{n}$ new people.

Recursive rules. A recursive rule gives the first term(s) and a rule for each later term in terms of earlier ones. An AP is $a_{n + 1} = a_{n} + d$ and a GP is $a_{n + 1} = r a_{n}$ . The Virahānka–Fibonacci sequence is $u_{1} = 1$ , $u_{2} = 2$ , $u_{n} = u_{n - 1} + u_{n - 2}$ , giving $1, 2, 3, 5, 8, 13, 21, 34, \dots$ . It counts the patterns of short beats (length 1) and long beats (length 2) with a total length of $n$ : the last beat is either short (leaving $n - 1$ ) or long (leaving $n - 2$ ). For total length $5$ there are $8$ patterns: $1$ with no long beat, $4$ with one long beat, $3$ with two. Starting with $1, 1$ gives the Fibonacci numbers, whose first ten terms $1, 1, 2, 3, 5, 8, 13, 21, 34, 55$ sum to $143$ , which is $144 - 1$ (the sum of the first $n$ is the term two places further on, minus $1$ ). The ratio of consecutive terms approaches about $1.618$ .

Tower of Hanoi. To move $n$ discs from one peg to another, never placing a larger disc on a smaller one, you must move the top $n - 1$ discs away, move the largest, then move the $n - 1$ discs back on top: $M_{n} = 2 M_{n - 1} + 1$ , with $M_{1} = 1$ . So $M_{n} = 1, 3, 7, 15, 31, 63, \dots = 2^{n} - 1$ . If each move takes 2 seconds, 6 discs need $63 \times 2 = 126$ seconds. The number of discs for which the moves first exceed $1000$ is $10$ ( $2^{10} - 1 = 1023$ ).

Fractals. The Sierpiński triangle: at each stage every shaded triangle is replaced by three smaller triangles (a central one is removed), so the number of triangles is $3^{n}$ and the shaded area is $(\frac{3}{4})^{n}$ of the original: at stage $4$ there are $81$ triangles and $\frac{81}{256}$ of the area. The Sierpiński carpet: each square is split into $9$ and the centre removed, so there are $8^{n}$ squares of side $\frac{1}{3 ^{n}}$ and the remaining area fraction is $(\frac{8}{9})^{n}$ . For an original side of $27$ cm, at stage $3$ there are $512$ squares of side $1$ cm each, a total area of $512$ cm $^{2}$ out of $729$ cm $^{2}$ .

Method. Decide AP, GP or recursive; write the first term and the rule; check the rule on the given terms; then compute the required term or total.

Geometric Progressions, Recursive Rules and Fractal Patterns: formulas to remember

GP nth term: $a_{n} = a r^{n - 1}$; $r$ = common ratio.
Common ratio: $r = \frac{a _{n + 1}}{a _{n}}$; Constant for a GP.
Recursive rules: $A P : a_{n + 1} = a_{n} + d; GP : a_{n + 1} = r a_{n}$; Plus the first term.
Virahānka–Fibonacci: $u_{n} = u_{n - 1} + u_{n - 2}, u_{1} = 1, u_{2} = 2$; Counts patterns of short and long beats.
Tower of Hanoi: $M_{n} = 2 M_{n - 1} + 1 = 2^{n} - 1$; Minimum number of moves.
Sierpiński patterns: $triangle: 3^{n}, (\frac{3}{4})^{n}; carpet: 8^{n}, (\frac{8}{9})^{n}$; Number of pieces and area fraction at stage $n$ .

How to solve Geometric Progressions, Recursive Rules and Fractal Patterns questions

Decide the type of sequence first; then use its rule.

1Compute successive differences and ratios to decide AP, GP or neither.
2For a GP, find $a$ and $r$ ; use $a_{n} = a r^{n - 1}$ , and for $n$ solve the power equation by comparing powers.
3For terms given far apart, divide to get $r^{k}$ and take the root.
4For three numbers in GP use $\frac{a}{r}$ , $a$ , $a r$ .
5For a recursive sequence, list terms one by one using the rule and check the pattern against any given term.
6For fractal and puzzle problems, find the pattern for stage $n$ and test it on the first three stages.

Geometric Progressions, Recursive Rules and Fractal Patterns in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Geometric progression

Suppose a message is forwarded in rounds: in round 1 the sender forwards it to 3 people, and in every later round each person who received it in the previous round forwards it to 3 new people.

How many people receive the message in round 5?

Try it first, then show the solutionSolution

Round 1: 3, round 2: $3 \times 3 = 9$ , round 3: $27$ . The numbers form a GP with $a = 3$ , $r = 3$ .
Round 5: $a_{5} = 3 \times 3^{4} = 243$ .

Answer: 243 people receive the message in round 5.

Memory hook: Multiply by the same number each time.

Tricks to remember Geometric Progressions, Recursive Rules and Fractal Patterns

Short memory aids. Say each one aloud once, then use it on a question the same day.

Difference or ratio
Constant difference: AP. Constant ratio: GP. Check both before choosing a formula.
Use it when: Identifying the sequence.
Each term is the sum of two before
Virahānka–Fibonacci: add the last two terms to get the next.
Use it when: Recursive rule questions.
Hanoi: double and add one
$M_{n} = 2 M_{n - 1} + 1$ : from 1, 3, 7, 15, 31, ...
Use it when: Tower of Hanoi.

Geometric Progressions, Recursive Rules and Fractal Patterns: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: GP term and position · 3 marks · Short answerBoard level

For the GP $3, 6, 12, \dots$ find the 8th term and the position of 768.

Try it first, then show the solutionSolution

$a = 3$ and $r = \frac{6}{3} = 2$ . (½)
$a_{8} = 3 \times 2^{7} = 3 \times 128 = 384$ . (1)
$3 \times 2^{n - 1} = 768$ gives $2^{n - 1} = 256 = 2^{8}$ , so $n - 1 = 8$ and $n = 9$ . (1½)

Answer: $a_{8} = 384$ and 768 is the 9th term.

Example 2: GP from two terms · 3 marks · Short answerBoard level

The 3rd term of a GP is 12 and the 6th term is 96. Find the common ratio, the first term and the first five terms.

Try it first, then show the solutionSolution

$a r^{2} = 12$ and $a r^{5} = 96$ ; dividing, $r^{3} = 8$ so $r = 2$ . (1½)
$a = \frac{12}{4} = 3$ . (½)
The terms are $3, 6, 12, 24, 48$ ; check $a_{6} = 3 \times 32 = 96$ . (1)

Answer: $r = 2$ , $a = 3$ ; the terms are $3, 6, 12, 24, 48$ .

Example 3: A recursive rule · 5 marks · Long answerHarder board level

The number of ways $u_{n}$ to build a pattern of total length $n$ using short beats (length 1) and long beats (length 2) satisfies $u_{n} = u_{n - 1} + u_{n - 2}$ , with $u_{1} = 1$ and $u_{2} = 2$ . Find $u_{3}$ to $u_{8}$ , justify the rule, and verify $u_{5}$ by counting.

Try it first, then show the solutionSolution

The last beat is short (leaving a pattern of length $n - 1$ ) or long (leaving length $n - 2$ ), so $u_{n} = u_{n - 1} + u_{n - 2}$ . (1)
$u_{3} = 3$ , $u_{4} = 5$ , $u_{5} = 8$ , $u_{6} = 13$ , $u_{7} = 21$ , $u_{8} = 34$ . (1½)
Count for total length 5 by the number of long beats: none gives $1$ pattern (11111); one long beat with three short ones: the long beat can be in any of 4 positions, giving $4$ ; two long beats with one short: 3 positions for the short one, giving $3$ . (1½)
Total $1 + 4 + 3 = 8 = u_{5}$ . (1)

Answer: $u_{3}, \dots, u_{8} = 3, 5, 8, 13, 21, 34$ ; $u_{5} = 8$ is confirmed by counting.

Example 4: Sierpiński triangle · 3 marks · Short answerHarder board level

In the Sierpiński triangle pattern, every shaded triangle is replaced at each stage by three triangles of half the side, with the central triangle removed. Find the number of shaded triangles and the fraction of the original area that remains at stage 4.

Try it first, then show the solutionSolution

Each stage multiplies the number of triangles by $3$ : stage 4 has $3^{4} = 81$ triangles. (1)
Each new triangle has $\frac{1}{4}$ of the area of the one it came from, so the area is multiplied by $\frac{3}{4}$ at each stage. (1)
Fraction remaining $= (\frac{3}{4})^{4} = \frac{81}{256}$ . (1)

Answer: 81 triangles; $\frac{81}{256}$ of the original area remains.

Example 5: Tower of Hanoi · 4 marks · Case-basedHOTS / competency

Case: in the Tower of Hanoi puzzle, $n$ discs are moved between three pegs, one at a time, never placing a larger disc on a smaller one. (i) Find the minimum number of moves for 4 discs. (ii) Show how the rule $M_{n} = 2 M_{n - 1} + 1$ gives the value for 5 discs. (iii) If each move takes 2 seconds, how long do 6 discs take? (iv) What is the least number of discs for which the minimum number of moves exceeds 1000?

Try it first, then show the solutionSolution

(i) $M_{n} = 2^{n} - 1$ , so $M_{4} = 15$ . (1)
(ii) $M_{5} = 2 \times 15 + 1 = 31 = 2^{5} - 1$ . (1)
(iii) $M_{6} = 63$ moves, so $63 \times 2 = 126$ seconds, that is 2 minutes 6 seconds. (1)
(iv) $2^{9} - 1 = 511$ and $2^{10} - 1 = 1023 > 1000$ , so 10 discs. (1)

Answer: (i) 15; (ii) 31; (iii) 126 s; (iv) 10 discs.

Where students lose marks in Geometric Progressions, Recursive Rules and Fractal Patterns

The slips examiners see most often, and what to write instead.

Mistake: Using n for the power
Writing $a r^{n}$ for the nth term.
Wrong working
$a_{5} = 2 \times 3^{5}$ .
Right working
$a_{5} = 2 \times 3^{4} = 162$ .
Do this: The exponent is $n - 1$ .
Usually costs: The whole answer.
Mistake: Calling a sequence a GP after checking one ratio
Not checking all consecutive ratios.
Wrong working
$\frac{4}{1} = 4$ for $1, 4, 9$ so it is a GP.
Right working
$\frac{9}{4} \neq = 4$ , so it is not a GP.
Do this: Check at least two ratios.
Usually costs: 1 mark.
Mistake: Hanoi count as 2^n
Forgetting the "minus 1".
Wrong working
$M_{4} = 2^{4} = 16$ .
Right working
$M_{4} = 2^{4} - 1 = 15$ .
Do this: Check with $M_{1} = 1$ .
Usually costs: 1 mark.

Exam tips

Name the type of sequence and justify it with the common difference or ratio.
Write the first few terms of a recursive sequence to check your rule.
For fractal problems, find the factor for number and for area at each stage.
State the final answer with its units in context problems.

Practice questions on Geometric Progressions, Recursive Rules and Fractal Patterns

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The 5th term of the GP $2, 6, 18, \dots$ is
(a) 54
(b) 162
(c) 486
(d) 108
Hint
$a_{5} = a r^{4}$ with $r = 3$ .
Answer
(b) 162. $a_{5} = 2 \times 3^{4} = 162$ .
Q2 · 3 marks · Short answerBoard level
A colony starts with 500 bacteria and the number doubles every hour. Find the number after 6 hours, and the number of hours after which the colony first exceeds 100000.
Hint
The number after $n$ hours is $500 \times 2^{n}$ .
Answer
After 6 hours: $500 \times 2^{6} = 32000$ . $500 \times 2^{n} > 100000$ means $2^{n} > 200$ ; $2^{7} = 128 < 200$ and $2^{8} = 256 > 200$ , so after 8 hours (128000 bacteria).
Q3 · 3 marks · Short answerHarder board level
Classify each sequence as AP, GP or neither, with reasons: (i) 2, 6, 18, 54 (ii) 5, 9, 13, 17 (iii) 1, 4, 9, 16.
Hint
Compute differences and ratios.
Answer
(i) Ratios $3, 3, 3$ : a GP with $r = 3$ . (ii) Differences $4, 4, 4$ : an AP with $d = 4$ . (iii) Differences $3, 5, 7$ are not constant and ratios $4, 2.25, 1.78$ are not constant, so neither.
Q4 · 5 marks · Long answerHarder board level
Three numbers in GP have sum 21 and product 216. Find the numbers and show that the two possible answers are the same GP read in opposite directions.
Hint
Write the numbers as $\frac{a}{r}$ , $a$ , $a r$ .
Answer
The product is $a^{3} = 216$ , so $a = 6$ . The sum: $6 (\frac{1}{r} + 1 + r) = 21$ , so $r + \frac{1}{r} = \frac{5}{2}$ and $2 r^{2} - 5 r + 2 = 0$ , giving $(2 r - 1) (r - 2) = 0$ , $r = 2$ or $r = \frac{1}{2}$ . For $r = 2$ : $3, 6, 12$ . For $r = \frac{1}{2}$ : $12, 6, 3$ . Check $3 + 6 + 12 = 21$ , $3 \times 6 \times 12 = 216$ . The second is the first in reverse (the ratios are reciprocals).
Q5 · 4 marks · Case-basedHOTS / competency
Case: the Sierpiński carpet starts with a square of side 27 cm. At each stage every shaded square is divided into 9 equal squares and the centre one is removed. (i) How many shaded squares are there at stage 3? (ii) What fraction of the original area remains at stage 2? (iii) What is the side of each smallest square at stage 3? (iv) What is the total shaded area at stage 3?
Hint
The number of squares is multiplied by 8 and the side divided by 3 at each stage.
Answer
(i) $8^{3} = 512$ . (ii) $(\frac{8}{9})^{2} = \frac{64}{81}$ . (iii) $\frac{27}{3 ^{3}} = 1$ cm. (iv) $512 \times 1^{2} = 512$ cm $^{2}$ , which is $\frac{512}{729}$ of the original $729$ cm $^{2}$ , matching $(\frac{8}{9})^{3}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.nth term of a GP?Show answer
$a r^{n - 1}$ .
2.Fibonacci recursion?Show answer
Each term is the sum of the previous two.
3.Minimum moves for n discs in Hanoi?Show answer
$2^{n} - 1$ .
4.Number of triangles at stage n of the Sierpiński triangle?Show answer
$3^{n}$ .
5.Area fraction remaining in the Sierpiński carpet at stage n?Show answer
$(\frac{8}{9})^{n}$ .

Geometric Progressions, Recursive Rules and Fractal Patterns: frequently asked questions

Can the common ratio be negative?

Yes. Then the signs alternate, like 2, -4, 8, -16, \dots.

Why is the Fibonacci rule called Virahānka–Fibonacci here?

Because the same sequence was described in India for counting patterns of short and long syllables before it became known in Europe.

Does the Sierpiński triangle ever have zero area?

The shaded area is \left(\frac{3}{4}\right)^n of the original, which gets smaller and smaller and tends to zero as n grows.