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Class 9 Maths Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions — Notes, Examples & Practice

Patterns let you predict: the next seat number, the next month's savings, the number of branches on a growing fractal. This chapter teaches two ways to describe a sequence — a direct (explicit) rule and a step-by-step (recursive) rule — and then studies the two most important families: arithmetic progressions, which grow by adding the same amount, and geometric progressions, which grow by multiplying by the same amount.

1Sequences: explicit and recursive rules

A sequence is an ordered list of numbers a1,a2,a3,…a_1, a_2, a_3, \dots; the subscript is the position nn and ana_n is the term at that position. Position and value are different things — the 4th term of 5, 9, 13, 17 is 17.

An explicit rule gives any term straight from nn, for example an=4n+1a_n = 4n + 1. You can jump to a100a_{100} in one step. A recursive rule gives each term from the previous ones and needs starting value(s): a1=5a_1 = 5, an=an−1+4a_n = a_{n-1} + 4. Both rules describe the same sequence here.

The Virahāṅka–Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, … has the recursive rule an=an−1+an−2a_n = a_{n-1} + a_{n-2}. The Indian scholar Virahāṅka found it (around the 7th century) while counting rhythms made of short and long syllables, centuries before Fibonacci in Europe.

Key formulas & definitions
  • Explicit: an=f(n)a_n = f(n) · Recursive: ana_n from an−1a_{n-1} (and earlier) + starting values
  • Virahāṅka–Fibonacci: a1=a2=1a_1 = a_2 = 1, an=an−1+an−2a_n = a_{n-1} + a_{n-2}
Worked example 1

Write the first four terms of (i) an=n2−1a_n = n^2 - 1, (ii) a1=3a_1 = 3, an=2an−1−1a_n = 2a_{n-1} - 1.

  1. (i) n=1,2,3,4n = 1, 2, 3, 4: 0, 3, 8, 15.
  2. (ii) a1=3a_1 = 3, a2=2(3)−1=5a_2 = 2(3) - 1 = 5, a3=2(5)−1=9a_3 = 2(5) - 1 = 9, a4=17a_4 = 17.

Answer: (i) 0, 3, 8, 15 (ii) 3, 5, 9, 17.

Common mistakes
  • Giving a recursive rule without a starting value — it then describes infinitely many sequences.
  • Confusing an−1a_{n-1} (the previous term) with an−1a_n - 1 (one less than this term).

2Arithmetic progressions

An arithmetic progression (AP) has a constant common difference d=an+1−and = a_{n+1} - a_n. Starting from the first term aa, each step adds dd, so after n−1n - 1 steps: an=a+(n−1)da_n = a + (n - 1)d.

d>0d > 0 gives an increasing AP, d<0d < 0 a decreasing one, d=0d = 0 a constant sequence. Plotted as points (n,an)(n, a_n), an AP lies on a straight line — it is linear growth (Chapter 2) in disguise.

To test whether a list is an AP, compute every consecutive difference in the same order (later minus earlier) and check they are all equal.

Key formulas & definitions
  • an=a+(n−1)da_n = a + (n - 1)d
  • d=a2−a1=a3−a2=…d = a_2 - a_1 = a_3 - a_2 = \dots
  • Number of terms: n=an−ad+1n = \dfrac{a_n - a}{d} + 1
Worked example 1

Find the 30th term of 7, 3, −1, −5, … and decide whether −81 is a term.

  1. a=7a = 7, d=3−7=−4d = 3 - 7 = -4.
  2. a30=7+29(−4)=7−116=−109a_{30} = 7 + 29(-4) = 7 - 116 = -109.
  3. For −81: 7+(n−1)(−4)=−81⇒(n−1)=22⇒n=237 + (n - 1)(-4) = -81 \Rightarrow (n - 1) = 22 \Rightarrow n = 23, a whole number.

Answer: a30=−109a_{30} = -109; −81 is the 23rd term.

Worked example 2

How many three-digit numbers are divisible by 7?

  1. Smallest: 105; largest: 994. AP with d=7d = 7.
  2. n=994−1057+1=127+1=128n = \dfrac{994 - 105}{7} + 1 = 127 + 1 = 128.

Answer: 128.

Trick / memory aid

Fence-post rule: from the 1st to the nnth term there are n−1n - 1 gaps, so you add dd only n−1n - 1 times.

Common mistakes
  • Using a+nda + nd for the nnth term.
  • Calculating dd as earlier − later and getting the sign wrong.

3Sum of the first n natural numbers

To add 1+2+⋯+n1 + 2 + \dots + n, write the sum twice — forwards and backwards — and add the columns: every column gives n+1n + 1, and there are nn columns. So twice the sum is n(n+1)n(n + 1).

A picture shows the same thing: a staircase of 1,2,…,n1, 2, \dots, n dots, together with a copy turned upside down, fills an n×(n+1)n \times (n + 1) rectangle. The numbers 1,3,6,10,15,…1, 3, 6, 10, 15, \dots are called triangular numbers.

The same pairing idea works for any AP: the sum of an AP is (number of terms) × (average of first and last term).

Key formulas & definitions
  • 1+2+3+⋯+n=n(n+1)21 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}
  • Sum of an AP =n2(first+last)= \dfrac{n}{2}(\text{first} + \text{last})
  • Sum of the first nn odd numbers =n2= n^2
Worked example 1

Find 1+2+⋯+1001 + 2 + \dots + 100 and the sum of all even numbers from 2 to 100.

  1. 100×1012=5050\dfrac{100 \times 101}{2} = 5050.
  2. 2+4+⋯+100=2(1+2+⋯+50)=2×50×512=25502 + 4 + \dots + 100 = 2(1 + 2 + \dots + 50) = 2 \times \dfrac{50 \times 51}{2} = 2550.

Answer: 5050 and 2550.

Common mistakes
  • Forgetting to divide by 2.

4Geometric progressions

A geometric progression (GP) has a constant non-zero common ratio r=an+1anr = \dfrac{a_{n+1}}{a_n}. Each step multiplies by rr, so an=arn−1a_n = a r^{n-1}.

If r>1r > 1 the terms grow faster and faster (doubling: 1, 2, 4, 8, …); if 0<r<10 < r < 1 they shrink towards 0 (halving); a negative rr makes the signs alternate. Unlike an AP, the points (n,an)(n, a_n) of a GP do not lie on a straight line.

The Tower of Hanoi puzzle is a classic: moving nn discs takes 2n−12^n - 1 moves at least, because each extra disc more than doubles the work.

Key formulas & definitions
  • an=arn−1a_n = a r^{n-1}
  • r=a2a1=a3a2=…r = \dfrac{a_2}{a_1} = \dfrac{a_3}{a_2} = \dots
Worked example 1

Find the 8th term of 3, 6, 12, 24, … and of 48, −24, 12, −6, …

  1. First: a=3a = 3, r=2r = 2: a8=3×27=384a_8 = 3 \times 2^7 = 384.
  2. Second: a=48a = 48, r=−12r = -\dfrac{1}{2}: a8=48×(−12)7=−48128=−38a_8 = 48 \times \left(-\dfrac{1}{2}\right)^7 = -\dfrac{48}{128} = -\dfrac{3}{8}.

Answer: 384 and −38-\dfrac{3}{8}.

Worked example 2

A ball is dropped from 16 m and each bounce reaches 34\dfrac{3}{4} of the previous height. How high is the 4th bounce?

  1. Bounce heights: 16×3416 \times \dfrac{3}{4}, 16×(34)216 \times \left(\dfrac{3}{4}\right)^2, …
  2. 4th bounce =16×(34)4=16×81256=5.0625= 16 \times \left(\dfrac{3}{4}\right)^4 = 16 \times \dfrac{81}{256} = 5.0625 m.

Answer: About 5.06 m.

Trick / memory aid

AP: same difference, add. GP: same ratio, multiply. Test a list by checking both: 2, 6, 18 has ratio 3 (GP), not a constant difference.

Common mistakes
  • Writing an=arna_n = a r^n — the power is n−1n - 1.
  • Deciding AP or GP from a graph's look instead of computing differences or ratios.

5Fractals and visualising growth

A fractal is a shape built by repeating the same step at smaller and smaller scales. Start with an equilateral triangle, join the midpoints of its sides and remove the middle triangle: you get 3 smaller triangles. Repeat on each: 9, then 27 … — the counts form a GP with ratio 3 (the Sierpiński triangle).

Each new triangle has 14\dfrac{1}{4} of the area of its parent, so the shaded area shrinks by a factor of 34\dfrac{3}{4} each stage: another GP.

Comparing an AP and a GP side by side — dots added in rows versus dots doubled — shows why multiplying eventually overtakes adding, however large the added amount.

Key formulas & definitions
  • Sierpiński: triangles at stage nn: 3n3^n; shaded area: A(34)nA \left(\dfrac{3}{4}\right)^n
  • GP with r>1r > 1 eventually beats any AP.
Worked example 1

Option A: get ₹1000 every day for 30 days. Option B: get ₹1 on day 1, ₹2 on day 2, ₹4 on day 3, doubling. On which day does B's daily amount first exceed A's?

  1. B's amount on day nn is 2n−12^{n-1}.
  2. 29=5122^9 = 512, 210=10242^{10} = 1024. So on day 11, B gives ₹1024 > ₹1000.

Answer: Day 11.

Common mistakes
  • Thinking area and count change by the same factor in a fractal.

Important questions

Try each question on paper first, then tap it to check the answer.

Basic Recall and direct use

Is 2, 5, 8, 11, … an AP? Find its 15th term.Answer
Yes, d=3d = 3. a15=2+14×3=44a_{15} = 2 + 14 \times 3 = 44.
Find the common ratio and the 6th term of 5, 15, 45, …Answer
r=3r = 3; a6=5×35=1215a_6 = 5 \times 3^5 = 1215.
Find the sum 1+2+⋯+401 + 2 + \dots + 40.Answer
40×412=820\dfrac{40 \times 41}{2} = 820.

Medium Two or three steps

The 5th term of an AP is 19 and the 12th term is 47. Find the first term and common difference.Answer
a+4d=19a + 4d = 19 and a+11d=47a + 11d = 47. Subtract: 7d=287d = 28, d=4d = 4, a=3a = 3.
Write both an explicit and a recursive rule for 4, 12, 36, 108, …Answer
Explicit: an=4⋅3n−1a_n = 4 \cdot 3^{n-1}. Recursive: a1=4a_1 = 4, an=3an−1a_n = 3a_{n-1}.
Rahul saves ₹50 in week 1, ₹60 in week 2, ₹70 in week 3, and so on. How much does he save in week 20, and in total over 20 weeks?Answer
AP with a=50a = 50, d=10d = 10. Week 20: 50+19×10=24050 + 19 \times 10 = 240. Total =202(50+240)=2900= \dfrac{20}{2}(50 + 240) = 2900 rupees.

Advanced Reasoning and HOTS

Which term of the GP 2, 6, 18, … is 1458?Answer
2⋅3n−1=1458⇒3n−1=729=36⇒n=72 \cdot 3^{n-1} = 1458 \Rightarrow 3^{n-1} = 729 = 3^6 \Rightarrow n = 7.
In how many ways can you climb a staircase of 6 steps if each move is 1 step or 2 steps? Explain the link with the Virahāṅka–Fibonacci sequence.Answer
Let wnw_n be the ways for nn steps. The last move is a 1-step (leaving n−1n - 1 steps) or a 2-step (leaving n−2n - 2), so wn=wn−1+wn−2w_n = w_{n-1} + w_{n-2} with w1=1w_1 = 1, w2=2w_2 = 2. Then w3=3w_3 = 3, w4=5w_4 = 5, w5=8w_5 = 8, w6=13w_6 = 13. Same idea as counting rhythms of short (1) and long (2) syllables.

Self-check quiz

Five quick questions. Tap an option to see if you are right.

The nnth term of an AP with first term aa and difference dd is
Which of these is a GP?
1+2+⋯+201 + 2 + \dots + 20 equals
The next term of 1, 1, 2, 3, 5, 8, … is
A recursive rule must always include

FAQ

What is the difference between an AP and a GP?

In an AP you add the same number each time (constant difference); in a GP you multiply by the same number each time (constant ratio).

Is every sequence an AP or a GP?

No. Square numbers 1, 4, 9, 16 and the Virahāṅka–Fibonacci numbers are neither. Always check differences and ratios.

Do I need to learn the sum of a GP?

For this chapter, focus on the nnth term of an AP and GP and the sum 1+2+⋯+n1 + 2 + \dots + n. Full sum formulas for APs come in Class 10.

Book references

  • NCERT Ganita Manjari Part I, Chapter 8: Predicting What Comes Next: Exploring Sequences and ProgressionsRead the chapter and do every exercise set first.
  • Mathematics for Class 9 — R.D. Sharma (Dhanpat Rai)Extra graded practice; its chapter numbers follow the old syllabus, so match by topic.
  • Mathematics for Class 9 — R.S. Aggarwal (Bharati Bhawan)Good for drill on basics; match by topic.
  • Exemplar Problems in Mathematics, Class IX — NCERTHarder, concept-testing questions (written for the old book).

These notes, examples and questions are written by PrepRung in our own words to help you study the NCERT Ganita Manjari book. NCERT and CBSE are not affiliated with PrepRung. Always study from your textbook too.