Sequences and Progressions teaches you to describe number patterns and predict what comes next. It starts with familiar sequences such as odd, square and triangular numbers, then introduces two ways to write a rule: the explicit rule, which gives any term from its position, and the recursive rule, which builds each term from earlier ones. It studies arithmetic progressions, geometric progressions and the sum of the first n natural numbers, with links to fractals and the Virahānka sequence. The nth term formula of an AP is the starting point for Class 10 Arithmetic Progressions.
Key Concepts
1. Sequences and Terms
A sequence is an ordered list of numbers where each number is a term of the sequence. The three dots … mean the sequence continues without end.
| Sequence | First six terms | Differences | Pattern |
|---|---|---|---|
| Natural numbers | 1, 2, 3, 4, 5, 6, … | 1, 1, 1, 1, 1 | Add 1 each time |
| Odd numbers | 1, 3, 5, 7, 9, 11, … | 2, 2, 2, 2, 2 | Add 2 each time |
| Triangular numbers | 1, 3, 6, 10, 15, 21, … | 2, 3, 4, 5, 6 | Sum of natural numbers up to that position |
| Square numbers | 1, 4, 9, 16, 25, 36, … | 3, 5, 7, 9, 11 | Sum of odd numbers up to that position |
Triangular numbers are sums: 3 = 1 + 2, 6 = 1 + 2 + 3, 10 = 1 + 2 + 3 + 4, and 15 = 1 + 2 + 3 + 4 + 5. Rows of 1, 2, 3, 4 and 5 dots form a triangle. Square numbers are sums of odd numbers: 4 = 1 + 3, 9 = 1 + 3 + 5, 16 = 1 + 3 + 5 + 7. Growing a 3 × 3 square of dots into a 4 × 4 square adds one row and one column sharing a corner, 4 + 4 − 1 = 7 dots, the next odd number.
Finite and infinite. 6, 12, 24, 48, 96 has five terms, so it is finite. The sequences in the table never end, so they are infinite.
Notation. t1 is the first term, t2 the second, and tn the term in position n. For the odd numbers, t4 = 7 means “the term in the fourth position is 7”. Use different letters (t, s, u) when a problem has more than one sequence.
Terms can be any real numbers. In 1, 1/2, 1/3, 1/4, … the terms are unit fractions that decrease. In −7, −3, 1, 5, 9, … the first term s1 is −7. The position n is always a non-negative integer, but the term itself can be negative, a fraction or any real number.
Not every sequence has a simple rule. The primes 2, 3, 5, 7, 11, 13, … have irregular gaps and no simple formula for the next term.
2. Explicit Rule for the nth Term
An explicit rule uses the position number n to calculate the nth term directly, without any earlier term.
Textbook example. un = 2n − 1 gives u1 = 1, u2 = 3, u3 = 5, so it is the explicit rule for the odd numbers. It jumps straight to far-off terms: the 53rd term is 2 × 53 − 1 = 105, the 108th is 215 and the 1170th is 2339.
Testing whether a number is a term. Set the rule equal to the number and solve for n. The number is a term only when n is a natural number, because n is a position.
- 2n − 1 = 137 gives n = 69, so 137 is the 69th odd number.
- For sn = 5n − 2 (terms 3, 8, 13, 18, 23, 28, …; 100th term 498): 5n − 2 = 308 gives 5n = 310, n = 62, so 308 is the 62nd term.
- 5n − 2 = 471 gives 5n = 473, n = 94.6. This is not a natural number, so 471 is not a term.
- The textbook first names 473 and then works 471. For 473: 5n = 475, n = 95, so 473 is the 95th term.
Worked illustration (tn = 3n − 7). First three terms: −4, −1, 2. The 12th, 18th and 50th terms are 29, 47 and 143. For 332: 3n = 339, n = 113, so 332 is the 113th term. For 557: 3n = 564, n = 188, a natural number, so 557 is a term (the 188th).
3. Recursive Rule for a Sequence
A recursive rule gives each term using earlier terms. It has two parts: the starting value or values and the rule for the next term.
For 1, 4, 7, 10, 13, … (explicit rule tn = 3n − 2), each term is 3 more than the previous one:
t1 = 1, tn = tn−1 + 3 for n ≥ 2
Without t1 = 1 the rule is incomplete, since tn = tn−1 + 3 also fits 5, 8, 11, ….
Example 3 of the textbook. u1 = 1, un = 2un−1 + 3 for n ≥ 2. Then u2 = 2 × 1 + 3 = 5, u3 = 2 × 5 + 3 = 13, u4 = 2 × 13 + 3 = 29. Continuing, u5 = 61, u6 = 125 and u7 = 253. The terms increase and jump from 125 to 253, so 133 is not a term.
Example 4 of the textbook. s1 = 3, sn = sn−1(sn−1 − 1) for n ≥ 2 gives s2 = 3 × 2 = 6, s3 = 6 × 5 = 30, s4 = 30 × 29 = 870.
The Virahānka-Fibonacci sequence. A recursive rule can use two or more earlier terms. With V1 = 1, V2 = 2 and Vn = Vn−1 + Vn−2 for n ≥ 3, we get V3 = 3, V4 = 5, V5 = 8, and the sequence 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …. Virahānka first wrote it down in his work Vṛttajātisamuchaya in the 7th century CE, in the context of Prakrit meter and poetry. Gopāla (c. 1135 CE) and Hemachandra (c. 1150 CE) studied it further, and Fibonacci (c. 1200 CE) later.
| Feature | Explicit rule | Recursive rule |
|---|---|---|
| Uses | The position n | Previous term or terms |
| Finding the 100th term | One substitution | Needs all 99 earlier terms |
| Odd numbers | tn = 2n − 1 | t1 = 1, tn = tn−1 + 2 |
4. Arithmetic Progressions (AP)
A growing pattern. In the textbook’s pattern of tiny squares, Stage 1 has 1 square and each new stage adds 4 squares at the corners: 1, 5, 9, 13, …. By Stage n, 4 has been added (n − 1) times, so tn = 1 + (n − 1) × 4 = 4n − 3. Stage 10 has 37 squares and Stage 20 has 77.
Definition. An arithmetic progression (AP) is a list of numbers in which each term after the first is obtained by adding a fixed number d, the common difference, to the previous term. The first term is a.
- General form: a, a + d, a + 2d, a + 3d, …, a + (n − 1)d
- nth term: tn = a + (n − 1)d
- Recursive rule: t1 = a, tn = tn−1 + d for n ≥ 2
d can be positive (1, 4, 7, …, d = 3), negative (11, 7, 3, −1, −5, …, d = −4) or zero (5, 5, 5, …). To check for an AP, find several differences t2 − t1, t3 − t2, …; they must all be equal. Natural numbers, odd numbers and −7, −3, 1, 5, 9 are APs. Triangular numbers, squares, unit fractions and the Virahānka sequence are not.
Worked illustrations. 2, 5, 8, 11, … has tn = 2 + 3(n − 1) = 3n − 1. −5, −1, 3, 7, … has tn = 4n − 9. 1/2, 5/2, 9/2, 13/2, … has d = 2 and tn = (4n − 3)/2.
Visualising an AP. Plot the pairs (stage, squares): (1, 1), (2, 5), (3, 9), (4, 13), (5, 17). The points lie on a straight line, because each step of 1 to the right raises the point by the same amount d. A negative d gives a line sloping down.
Taxi fare (Example 5 of the textbook). A taxi charges a booking fee of ₹200 plus ₹40 per km. The fares after 1, 2, 3 km are ₹240, ₹280, ₹320, an AP with a = 240, d = 40. The nth term is 240 + (n − 1) × 40 = 200 + 40n. For 10 km the fare is ₹600.
5. Sum of the First n Natural Numbers
The reversal trick. Write S = 1 + 2 + … + 10 and, below it, S = 10 + 9 + … + 1. Each vertical pair adds to 11, so 2S = 11 × 10 = 110 and S = 55. For 1 + 2 + … + n, each of the n pairs adds to n + 1, so 2S = n(n + 1) and
Sn = 1 + 2 + 3 + … + n = n(n + 1)/2
The picture. Draw 1 + 2 + … + 6 as a staircase of circles. Two copies, one turned upside down, fit together along a zigzag line into a 7 × 6 rectangle. So 2 × (1 + 2 + … + 6) = 7 × 6, and in general 2 × (1 + 2 + … + n) = n × (n + 1).
History. The first known written mention of this result is in Āryabhaṭa’s Āryabhaṭīya, Chapter 2, Verse 19. Its second method: take the sum of the first and last terms, divide by two (the average), and multiply by the number of terms.
Sums that do not start at 1. 25 + 26 + … + 58 = S58 − S24 = (58 × 59)/2 − (24 × 25)/2 = 1711 − 300 = 1411. Subtract S24, not S25, so that 25 stays in the sum. Āryabhaṭa’s rule agrees: 34 terms, average (25 + 58)/2 = 41.5, and 41.5 × 34 = 1411.
Triangular numbers. The nth triangular number is Sn, so the 10th, 17th and 80th are 55, 153 and 3240.
6. Geometric Progressions (GP)
A doubling pattern. In another textbook pattern the number of green squares doubles at each stage: 3, 6, 12, 24, …. Written as 3, 3 × 2, 3 × 22, 3 × 23, this gives tn = 3 × 2n−1, with recursive rule t1 = 3, tn = 2tn−1 for n ≥ 2. By Stage 20 there are 3 × 219 = 1572864 squares, against 77 for the AP pattern.
Definition. A geometric progression (GP) is a list of numbers in which each term after the first is obtained by multiplying the previous term by a fixed number r, the common ratio.
- General form: a, ar, ar2, ar3, …, arn−1
- nth term: tn = arn−1
- Recursive rule: t1 = a, tn = r × tn−1 for n ≥ 2
To check for a GP, divide each term by the one before; the ratios must all be equal. 1, 2, 4, 8, … has r = 2; 1, 3, 9, 27, … has r = 3; 1, −1, 1, −1, … has r = −1, and a negative ratio makes the signs alternate.
Example 9 of the textbook. For 5, 15/4, 45/16, 135/64, …: 15/4 ÷ 5 = 3/4, 45/16 ÷ 15/4 = 45/16 × 4/15 = 3/4, and 135/64 ÷ 45/16 = 3/4. So it is a GP with a = 5, r = 3/4 and tn = 5 × (3/4)n−1.
Worked illustrations. 2, 10, 50, 250, … has tn = 2 × 5n−1. 3, −3/2, 3/4, −3/8, … has r = −1/2 and tn = 3 × (−1/2)n−1. For 3, 30, 300, … (tn = 3 × 10n−1) the recursive rule is t1 = 3, tn = 10tn−1.
Visualising a GP. The points (1, 3), (2, 6), (3, 12), (4, 24), (5, 48) do not lie on a straight line: the rise at each step keeps growing, so the points bend upwards more and more steeply. When r lies between 0 and 1, the points fall and get closer and closer to 0.
| Feature | AP | GP |
|---|---|---|
| Next term | Add d | Multiply by r |
| Test | Equal differences | Equal ratios |
| nth term | a + (n − 1)d | arn−1 |
| Graph of (n, tn) | Straight line | Curve |
Bouncing ball (Example 10 of the textbook). A ball dropped from 24 feet rises to 3/4 of its previous height after each bounce: 18, 13.5, 10.125, … ft, a GP with a = 18 and r = 0.75. The seventh bounce (about 3.20 ft) is the first below one sixth of 24 ft, which is 4 ft; the sixth reaches about 4.27 ft.
7. GPs in Fractals, and the Tower of Hanoi
The Sierpiński triangle. Stage 0 is a paper equilateral triangle. Join the midpoints of its sides to make four smaller equilateral triangles and remove the central one: this is Stage 1, with three black triangles. Repeat on every black triangle to get Stage 2, then Stage 3, and so on forever. The pattern is named after the Polish mathematician Wacław Sierpiński (1882 to 1969) and is a fractal.
Each black triangle becomes three, so the count is multiplied by 3 at every stage. Each stage keeps 3 of 4 equal parts, so the black area (Stage 0 area = 1 square unit) is multiplied by 3/4.
| Stage n | 0 | 1 | 2 | 3 | 4 | 5 | n |
|---|---|---|---|---|---|---|---|
| Black triangles | 1 | 3 | 9 | 27 | 81 | 243 | 3n |
| Black area | 1 | 3/4 | 9/16 | 27/64 | 81/256 | 243/1024 | (3/4)n |
With stages counted from 0, the recursive rules are t0 = 1, tn = 3tn−1 and s0 = 1, sn = (3/4)sn−1 for n ≥ 1. The textbook prints them starting from t1 = 1 and s1 = 1; that labels Stage 0 as term 1, which would make the count 3n−1. Keep the subscript equal to the stage number so the rule matches 3n. The count grows very quickly (r = 3), while the area shrinks towards 0 (r = 3/4).
Fractals are shapes that repeat themselves at different scales, like tree branches, cauliflower, snowflakes and coastlines.
The Tower of Hanoi. The CBSE Class IX course structure also lists this puzzle for the chapter. A stack of n discs of different sizes sits on one of three pegs, largest at the bottom. Move the whole stack to another peg, one top disc at a time, never placing a larger disc on a smaller one. To move n discs, move the top n − 1 to the spare peg, move the largest disc, then move the n − 1 discs onto the largest. So the least number of moves obeys M1 = 1, Mn = 2Mn−1 + 1, giving 1, 3, 7, 15, 31, …, and the explicit rule Mn = 2n − 1. This is one less than the GP 2, 4, 8, 16, 32, ….
Formula and Theorem Sheet
| Topic | Formula or rule | Notes |
|---|---|---|
| Triangular numbers | tn = n(n + 1)/2 | Recursive: t1 = 1, tn = tn−1 + n |
| AP, nth term | tn = a + (n − 1)d | Graph of (n, tn) is a straight line |
| AP, recursive | t1 = a, tn = tn−1 + d, n ≥ 2 | State the first term |
| AP, number of terms from a to l | n = (l − a)/d + 1 | From a + (n − 1)d = l |
| Sum of first n natural numbers | Sn = n(n + 1)/2 | Proof: pair the sum with its reverse |
| Sum from m to n | Sn − Sm−1 | Subtract up to m − 1, not m |
| GP, nth term | tn = arn−1 | Graph of (n, tn) is a curve |
| GP, recursive | t1 = a, tn = r × tn−1, n ≥ 2 | r < 0 makes signs alternate |
| Virahānka sequence | V1 = 1, V2 = 2, Vn = Vn−1 + Vn−2, n ≥ 3 | 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 |
| Sierpiński triangle | Triangles 3n, area (3/4)n | Stage 0 = one triangle of area 1 |
| Sierpiński square carpet | Squares 8n, area (8/9)n | Stage 0 = one square of area 1 |
| Tower of Hanoi, n discs | M1 = 1, Mn = 2Mn−1 + 1; Mn = 2n − 1 | Least number of moves |
Read the rest of the chapter →Hide the rest ↑
Important Definitions
- Sequence: an ordered list of numbers. Each number in the list is called a term.
- Finite and infinite sequences: a finite sequence has a last term (6, 12, 24, 48, 96); an infinite one continues without end (1, 3, 5, 7, …).
- Explicit formula: a rule that uses the term’s position number, n, to calculate the term’s value, such as tn = 2n − 1.
- Recursive formula: a rule that gives the value of a term using the values of previous terms, together with the starting term or terms, such as t1 = 1, tn = tn−1 + 3 for n ≥ 2.
- Arithmetic progression (AP): a list of numbers in which each term after the first is obtained by adding a fixed number d to the previous term.
- Common difference (d): the fixed number added at each step of an AP; d = tn − tn−1. It can be positive, negative or zero.
- Geometric progression (GP): a list of numbers in which each term after the first is obtained by multiplying the previous term by a fixed number.
- Common ratio (r): the constant factor of a GP; r = tn ÷ tn−1.
- Fractal: a shape or pattern that repeats itself at different scales, so a zoomed-in part looks similar to the whole.
Solved Examples (NCERT-Based)
Example 1: Terms from an explicit rule (Exercise Set 8.1, Q1)
Question: Find the first five terms of the sequence in which the nth term is given by (i) tn = 3n − 4, (ii) tn = 2 − 5n, and (iii) tn = n2 − 2n + 3 for n ≥ 1.
Solution: Substitute n = 1, 2, 3, 4, 5.
- (i) 3(1) − 4 = −1, 3(2) − 4 = 2, 3(3) − 4 = 5, 3(4) − 4 = 8, 3(5) − 4 = 11. Terms: −1, 2, 5, 8, 11.
- (ii) 2 − 5 = −3, 2 − 10 = −8, 2 − 15 = −13, 2 − 20 = −18, 2 − 25 = −23. Terms: −3, −8, −13, −18, −23.
- (iii) 1 − 2 + 3 = 2, 4 − 4 + 3 = 3, 9 − 6 + 3 = 6, 16 − 8 + 3 = 11, 25 − 10 + 3 = 18. Terms: 2, 3, 6, 11, 18.
Example 2: Checking whether numbers are terms (Exercise Set 8.1, Q3)
Question: Determine whether 97 and 172 are terms of the sequence tn = 5n − 3 for n ≥ 1.
Solution: For 97: 5n − 3 = 97, so 5n = 100 and n = 20. Since 20 is a natural number, 97 is a term (the 20th).
For 172: 5n − 3 = 172, so 5n = 175 and n = 35. Since 35 is a natural number, 172 is a term (the 35th).
Check: t20 = 100 − 3 = 97 and t35 = 175 − 3 = 172. Both numbers are terms.
Example 3: From a recursive rule to a term number (Exercise Set 8.1, Q5)
Question: A sequence is given by the recursive rule t1 = −5, tn+1 = tn + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
Solution: t1 = −5, t2 = −5 + 3 = −2, t3 = −2 + 3 = 1, t4 = 1 + 3 = 4, t5 = 4 + 3 = 7. The first five terms are −5, −2, 1, 4, 7.
Each term is 3 more than the previous one, so this is an AP with a = −5 and d = 3. Its explicit rule is tn = −5 + (n − 1) × 3 = 3n − 8.
Solve 3n − 8 = 52: 3n = 60, n = 20. So 52 is a term: the 20th term.
Example 4: A decreasing AP and the number 0 (Exercise Set 8.2, Q2)
Question: Which term of the AP : 21, 18, 15, … is −81? Also, is 0 a term of this AP? Give reasons for your answer.
Solution: a = 21, d = 18 − 21 = −3. So tn = 21 + (n − 1)(−3) = 24 − 3n.
For −81: 24 − 3n = −81, so 3n = 105 and n = 35. So −81 is the 35th term.
For 0: 24 − 3n = 0, so 3n = 24 and n = 8. Since 8 is a natural number, 0 is a term: the 8th term.
Example 5: Finding a term from two given terms (Exercise Set 8.2, Q4)
Question: An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Solution: The last term is the 50th term. So
- t3 = a + 2d = 12 … (1)
- t50 = a + 49d = 106 … (2)
Subtract (1) from (2): 47d = 94, so d = 2. Put d = 2 in (1): a + 4 = 12, so a = 8.
t29 = a + 28d = 8 + 28 × 2 = 8 + 56 = 64. The 29th term is 64.
Example 6: Two-digit multiples of 3 and their sum (Exercise Set 8.2, Q5)
Question: How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Solution: The smallest 2-digit multiple of 3 is 12 and the largest is 99. They form the AP 12, 15, 18, …, 99 with a = 12 and d = 3.
Number of terms: 12 + (n − 1) × 3 = 99, so (n − 1) × 3 = 87, n − 1 = 29 and n = 30. There are 30 such numbers.
Sum: every term is 3 times a natural number from 4 to 33. So the sum is 3 × (4 + 5 + … + 33) = 3 × (S33 − S3).
S33 = 33 × 34/2 = 561 and S3 = 6, so the sum = 3 × (561 − 6) = 3 × 555 = 1665.
Example 7: Salary with a fixed increment (Exercise Set 8.2, Q6)
Question: Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?
Solution: His yearly salaries form an AP: 5,00,000, 5,20,000, 5,40,000, … with a = 5,00,000 and d = 20,000. His salary in the nth year is tn = 5,00,000 + (n − 1) × 20,000.
Set tn = 7,00,000: (n − 1) × 20,000 = 2,00,000, so n − 1 = 10 and n = 11.
His salary is ₹7,00,000 in the 11th year of work, that is, after 10 years (10 increments of ₹20,000).
Example 8: A GP term from another term (Exercise Set 8.3, Q1)
Question: Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Solution: t8 = ar7 = a × 27 = 128a = 192, so a = 192/128 = 3/2.
t12 = ar11 = (3/2) × 211 = (3/2) × 2048 = 3072.
Shortcut: going from the 8th to the 12th term means multiplying by r four times, so t12 = 192 × 24 = 192 × 16 = 3072.
Example 9: A recursive rule that hides a GP (Exercise Set 8.3, Q3)
Question: A sequence is given by the recursive rule t1 = 2, tn+1 = 3tn − 2 for n ≥ 1. Which term of the sequence is 730?
Solution: Generate the terms:
- t2 = 3 × 2 − 2 = 4
- t3 = 3 × 4 − 2 = 10
- t4 = 3 × 10 − 2 = 28
- t5 = 3 × 28 − 2 = 82
- t6 = 3 × 82 − 2 = 244
- t7 = 3 × 244 − 2 = 730
So 730 is the 7th term.
Pattern behind it: subtract 1 from each term: 1, 3, 9, 27, 81, 243, 729. These are powers of 3, so tn = 3n−1 + 1. Then 3n−1 + 1 = 730 gives 3n−1 = 729 = 36, so n − 1 = 6 and n = 7, the same answer.
Example 10: The bouncing ball (Exercise Set 8.3, Q5)
Question: A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way, each time rising to 60% of the previous height. (i) What height does the ball reach after the 5th bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
Solution (i): The heights after each bounce form a GP with first term 80 × 0.6 = 48 and r = 0.6: 48, 28.8, 17.28, 10.368, 6.2208, …. The height after the nth bounce is 80 × (0.6)n, so after the 5th bounce it is 80 × 0.07776 = 6.2208 m.
Solution (ii): The ball falls 80 m before the 1st hit. Each of bounces 1 to 5 is a rise and an equal fall, and the fall after the 5th bounce ends in the 6th hit.
- First fall: 80 m
- Bounces 1 to 5, up and down: 2 × (48 + 28.8 + 17.28 + 10.368 + 6.2208) = 2 × 110.6688 = 221.3376 m
Total vertical distance = 80 + 221.3376 = 301.3376 m, about 301.34 m.
Example 11: Two conditions on an AP (End-of-Chapter Exercises, Q8)
Question: The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution: t4 + t8 = (a + 3d) + (a + 7d) = 2a + 10d = 24 … (1)
t6 + t10 = (a + 5d) + (a + 9d) = 2a + 14d = 44 … (2)
Subtract (1) from (2): 4d = 20, so d = 5. From (1): 2a + 50 = 24, so 2a = −26 and a = −13.
The first three terms are −13, −8, −3. Check: t4 = 2, t8 = 22, sum 24; t6 = 12, t10 = 32, sum 44.
Competency-Based Questions (with answers)
1. Case-based: Seats in the school auditorium
A school auditorium has 15 rows of seats. The first row has 20 seats and each row has 3 more seats than the row in front of it.
(a) How many seats are in the 10th row? The rows form an AP with a = 20, d = 3. t10 = 20 + 9 × 3 = 47 seats.
(b) Which row has 59 seats? 20 + (n − 1) × 3 = 59, so (n − 1) × 3 = 39 and n = 14. The 14th row.
(c) How many seats are there in all? Row n has 20 + 3(n − 1) seats. Adding over 15 rows: 15 × 20 + 3 × (0 + 1 + 2 + … + 14) = 300 + 3 × S14 = 300 + 3 × (14 × 15/2) = 300 + 3 × 105 = 615 seats.
2. Case-based: A message chain
At 9:00 a student sends a class notice to 2 friends. Every 10 minutes, each person who received the notice in the previous round sends it to 2 new people who have not received it.
(a) How many new people receive the notice in round 5? Round 1: 2, round 2: 4, round 3: 8. This is a GP with a = 2, r = 2, so round n reaches 2 × 2n−1 = 2n people. Round 5: 25 = 32.
(b) In which round do more than 1000 new people receive it for the first time? 29 = 512 and 210 = 1024. Round 10 is the first round above 1000.
(c) How many people have received the notice after 5 rounds, not counting the sender? 2 + 4 + 8 + 16 + 32 = 62.
3. Source-based: Āryabhaṭa’s rule
The first known written mention of the sum of natural numbers is in Āryabhaṭa’s Āryabhaṭīya, Chapter 2, Verse 19. The second method in the verse says the sum can be found by taking the sum of the first and last terms, dividing by two, and multiplying by the number of terms.
(a) Use the rule to find 1 + 2 + … + 100. (1 + 100)/2 × 100 = 50.5 × 100 = 5050.
(b) Use it to find 2 + 4 + 6 + … + 100. There are 50 terms, so (2 + 100)/2 × 50 = 51 × 50 = 2550.
4. Source-based: Virahānka and poetic rhythm
The textbook states that Virahānka discovered the sequence 1, 2, 3, 5, 8, … in the context of Prakrit meter and poetry. In such meters a line is built from short syllables (1 beat) and long syllables (2 beats).
(a) List all rhythms that fill exactly 3 beats. Short-short-short, short-long, long-short: 3 rhythms.
(b) Explain why the number of rhythms for n beats equals the number for (n − 1) beats plus the number for (n − 2) beats. Every n-beat rhythm ends with either a short syllable or a long one. If it ends with a short, the rest fills n − 1 beats; if it ends with a long, the rest fills n − 2 beats. So the counts add, which is the rule Vn = Vn−1 + Vn−2.
(c) How many rhythms fill 6 beats? The counts for 1, 2, 3, 4, 5, 6 beats are 1, 2, 3, 5, 8, 13. So 13 rhythms.
5. Assertion-Reason
Assertion (A): 471 is not a term of the sequence sn = 5n − 2.
Reason (R): The position n of a term must be a natural number.
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). 5n − 2 = 471 gives n = 94.6, which is not a natural number, so there is no such position. R is exactly why A holds.
6. Assertion-Reason
Assertion (A): The sequence 1, 4, 9, 16, 25, … is an arithmetic progression.
Reason (R): Every term of the sequence is a perfect square.
Answer: (d). A is false: the differences 3, 5, 7, 9 are not constant. R is true, but being perfect squares has nothing to do with being an AP.
7. Assertion-Reason
Assertion (A): In the Sierpiński triangle, the number of black triangles increases from stage to stage while the black area decreases.
Reason (R): From one stage to the next, the number of black triangles is multiplied by 3 and the black area is multiplied by 3/4.
Answer: (a). Multiplying by 3 (greater than 1) makes the count grow; multiplying by 3/4 (between 0 and 1) makes the area shrink. R explains A.
8. Error analysis
A student writes: “The 20th term of the AP 7, 11, 15, … is 7 + 20 × 4 = 87.”
(a) Find the error. The formula is tn = a + (n − 1)d. To reach the 20th term from the first, d is added 19 times, not 20.
(b) Give the correct answer. t20 = 7 + 19 × 4 = 7 + 76 = 83.
Important Questions for Exams
1-Mark Questions
- Find the common difference of the AP 11, 8, 5, 2, …. d = 8 − 11 = −3.
- Is 1, −1, 1, −1, 1, … a geometric progression? If so, what is the common ratio? Yes. Each term is −1 times the previous one, so r = −1.
- Write the 7th triangular number. 7 × 8/2 = 28.
- Write the nth term of the AP 2, 5, 8, 11, …. tn = 2 + (n − 1) × 3 = 3n − 1.
- Write the next two terms of 1, 2, 3, 5, 8, 13, 21, 34, …. 21 + 34 = 55 and 34 + 55 = 89.
2-Mark Questions
- Find the 10th and 26th terms of the AP: 3, 8, 13, 18, …. (Exercise Set 8.2, Q1) a = 3, d = 5, tn = 3 + 5(n − 1) = 5n − 2. t10 = 48 and t26 = 128.
- Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP. (Exercise Set 8.2, Q3) a = 11, d = −3, tn = 11 + (n − 1)(−3) = 14 − 3n. Recursive rule: t1 = 11, tn = tn−1 − 3 for n ≥ 2.
- Which term of the sequence tn = 5n − 3 for n ≥ 1 is 607? (Exercise Set 8.1, Q4) 5n − 3 = 607, so 5n = 610 and n = 122. It is the 122nd term.
- Find the 10th and nth terms of the GP: 5, 25, 125, …. (Exercise Set 8.3, Q2) a = 5, r = 5, so tn = 5 × 5n−1 = 5n. t10 = 510 = 9765625.
- A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all? (Exercise Set 8.2, Q7) 1 + 2 + … + 25 = S25 = 25 × 26/2 = 325 marbles.
3-Mark Questions
- Find the 31st term of an AP whose 11th term is 38 and 16th term is 73. (End-of-Chapter Exercises, Q1) a + 10d = 38 and a + 15d = 73. Subtracting, 5d = 35, so d = 7 and a = 38 − 70 = −32. t31 = a + 30d = −32 + 210 = 178.
- How many three-digit numbers are divisible by 7? (End-of-Chapter Exercises, Q3) The smallest is 105 (7 × 15) and the largest is 994 (7 × 142). They form an AP with a = 105, d = 7. 105 + (n − 1) × 7 = 994 gives (n − 1) × 7 = 889, n − 1 = 127, n = 128. There are 128 such numbers.
- Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000. (End-of-Chapter Exercises, Q9) We need n(n + 1)/2 > 1000, that is, n(n + 1) > 2000. For n = 44: 44 × 45 = 1980, so S44 = 990, not enough. For n = 45: 45 × 46 = 2070, so S45 = 1035 > 1000. The smallest n is 45.
- Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term. (End-of-Chapter Exercises, Q10) a = 2, r = 4, tn = 2 × 4n−1. 2 × 4n−1 = 131072 gives 4n−1 = 65536 = 48, so n = 9. It is the 9th term. Recursive formula: t1 = 2, tn = 4tn−1 for n ≥ 2.
5-Mark Questions
- Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term. (End-of-Chapter Exercises, Q5) Let the terms be a, ar, ar2, …. The fifth term is 4 times the third: ar4 = 4ar2. Since a ≠ 0 and r ≠ 0, divide by ar2: r2 = 4, so r = 2 or r = −2. Sum of first two terms: a + ar = a(1 + r) = −4. If r = 2: 3a = −4, a = −4/3, giving the GP −4/3, −8/3, −16/3, −32/3, …. If r = −2: −a = −4, a = 4, giving the GP 4, −8, 16, −32, 64, …. Check the second: 4 + (−8) = −4 and the fifth term 64 = 4 × 16. Both GPs satisfy the conditions.
- Based on Exercise Set 8.3, Q7: the Sierpiński square carpet. Stage 0 is a square of area 1 square unit. For Stage 1, each side is trisected, the points of trisection of opposite sides are joined to make nine smaller squares, and the centre square is removed, leaving 8. The process is repeated on every remaining square. Find the number of squares and the area at Stages 0 to 5, the explicit and recursive formulas for each, and what happens to the area as n increases.
Answer: Each square becomes 8 at the next stage, so the counts are 1, 8, 64, 512, 4096, 32768. Explicit: tn = 8n; recursive: t0 = 1, tn = 8tn−1 for n ≥ 1. Each stage keeps 8 of 9 equal parts, so the areas are 1, 8/9, 64/81, 512/729, 4096/6561, 32768/59049. Explicit: sn = (8/9)n; recursive: s0 = 1, sn = (8/9) × sn−1 for n ≥ 1. Since 8/9 lies between 0 and 1, the area keeps shrinking towards 0 while the number of squares grows very fast.
Common Mistakes and Examiner Tips
- Writing a + nd instead of a + (n − 1)d. The first term already sits at position 1, so d is added only n − 1 times. Test your formula on n = 1: it must give a.
- Accepting a decimal position. If solving tn = x gives n = 94.6 or n = −3, the number is not a term. Write the reason: “n is not a natural number”.
- A recursive rule without its starting term. tn = tn−1 + 3 fits infinitely many sequences. Always write t1 = … and the range, for example n ≥ 2.
- Getting the sign of d wrong in a decreasing AP. For 21, 18, 15, …, d = 18 − 21 = −3. Subtract the earlier term from the later term, never the other way.
- Calling a sequence an AP after checking one difference. 1, 3, 6, 10 has 3 − 1 = 2 but 6 − 3 = 3. Check at least three differences, and for a GP check at least three ratios.
- Writing arn for the nth term of a GP. The power of r is one less than the position: tn = arn−1. Check: t1 = ar0 = a.
- Subtracting the wrong partial sum. 25 + … + 58 = S58 − S24. Subtracting S25 drops 25 from the answer.
- Counting terms as (last − first)/d. From 12 to 99 in steps of 3 there are (99 − 12)/3 + 1 = 30 terms. Forgetting the +1 loses one term.
- Using the wrong limits in “between” and “three-digit” questions. For multiples of 4 between 10 and 250, the first is 12 and the last is 248, giving 60 multiples. For three-digit multiples of 7, start at 105, not 100 or 7.
- Stage 0 versus Stage 1 in fractal questions. In the Sierpiński triangle the stages start at 0, so Stage n has 3n triangles, not 3n−1. Read which stage the question calls the start.
Quick Revision Points
- A sequence is an ordered list of numbers; each number is a term.
- Explicit rule: term from position n. Recursive rule: term from earlier terms plus a starting value.
- To test if x is a term, solve tn = x; x is a term only if n is a natural number.
- Virahānka sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89; each term is the sum of the previous two; 7th century CE.
- AP: add a fixed d. tn = a + (n − 1)d; recursive t1 = a, tn = tn−1 + d.
- Plotting (n, tn) for an AP gives points on a straight line.
- Number of terms from a to l: (l − a)/d + 1.
- Sn = 1 + 2 + … + n = n(n + 1)/2, proved by adding the sum to its reverse.
- Āryabhaṭa (Āryabhaṭīya, Chapter 2, Verse 19): sum = average of first and last × number of terms.
- Sum from m to n = Sn − Sm−1.
- GP: multiply by a fixed r. tn = arn−1; recursive t1 = a, tn = r × tn−1.
- r > 1: terms grow fast; 0 < r < 1: terms shrink towards 0; r < 0: signs alternate.
- Plotting (n, tn) for a GP gives points on a curve, not a line.
- Sierpiński triangle: 3n triangles, area (3/4)n. Square carpet: 8n squares, area (8/9)n.
- Tower of Hanoi with n discs: Mn = 2Mn−1 + 1, least moves 2n − 1.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, Sequences and Progressions is one of four chapters in Unit II, Algebra, which carries 20 marks out of the 80-mark theory paper and is allotted 66 teaching periods in total. The course structure does not split the 20 marks chapter by chapter. The remaining 20 marks of the year are internal assessment (pen-paper tests and multiple assessment 10, portfolio 5, lab practical 5).
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra (includes Sequences and Progressions) | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The course structure asks students to find terms from explicit and recursive rules, identify and visualise APs and GPs, analyse fractals using GPs and solve the Tower of Hanoi puzzle. School papers test these through membership checks, nth-term questions, word problems and case-based questions on growing patterns or fractals. Class 10 Arithmetic Progressions starts from the nth term formula tn = a + (n − 1)d and the pairing idea behind Sn = n(n + 1)/2, so practise both until they are automatic.
Class 9 Maths · Chapter 8 – swipe through all 9 cards to understand the whole chapter.
Sequences and terms
A sequence is an ordered list of numbers; each number is a term, named by its position.
6, 12, 24, 48, 96 is finite; 1, 3, 5, 7, … is infinite.
- tₙ is the term in position n
- Terms can be negative or fractions
- Use different letters for different sequences
Jump to any term
An explicit rule uses the position n to calculate the term directly.
The 1170th odd number is 2 × 1170 − 1 = 2339.
- Solve tₙ = x to test if x is a term
- 137 is the 69th odd number
- n must come out as a natural number
Build from earlier terms
A recursive rule gives each term from the term or terms before it, plus a starting value.
u1 = 1, uₙ = 2uₙ₋1 + 3 gives 1, 5, 13, 29, 61, …
- Always state the first term
- Needs all earlier terms to reach a far term
- Can use two or more previous terms
Virahānka sequence
Each term after the second is the sum of the previous two.
Studied by Virahānka in the 7th century CE for Prakrit meter.
- 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
- Later studied by Gopāla and Hemachandra
- Fibonacci studied it around 1200 CE
Arithmetic progression
Each term after the first is obtained by adding a fixed number d.
Taxi fare 240, 280, 320, … gives tₙ = 200 + 40n.
- Test: all differences equal
- d can be negative, e.g. 11, 7, 3, −1
- Recursive: tₙ = tₙ₋1 + d
AP line, GP curve
Plot (n, tₙ): AP points lie on a straight line, GP points on a curve.
(1, 1), (2, 5), (3, 9), (4, 13) lie on one line; (1, 3), (2, 6), (3, 12) do not.
- AP rises by d at each step
- GP gaps grow when r is above 1
- GP terms shrink towards 0 when r is between 0 and 1
Sum of natural numbers
Add the sum to its reverse: n pairs, each adding to n + 1.
25 + 26 + … + 58 = S58 − S24 = 1711 − 300 = 1411.
- S10 = 55, S100 = 5050
- Also the nth triangular number
- Āryabhaṭīya, Chapter 2, Verse 19
Geometric progression
Each term after the first is obtained by multiplying by a fixed number r.
5, 15/4, 45/16, … is a GP with a = 5, r = 3/4.
- Test: all ratios equal
- r = −1 gives 1, −1, 1, −1, …
- Recursive: tₙ = r × tₙ₋1
Sierpiński triangle
Remove the middle triangle again and again; counts and areas form GPs.
Square carpet: 8ⁿ squares and area (8/9)ⁿ.
- Stages 0 to 5: 1, 3, 9, 27, 81, 243
- Area shrinks towards 0
- Fractals repeat at every scale
📝 Practice Sequences and Progressions - 10 questions
Concept-check questions · with answers & explanations
Start →Close ✕
Why B: computing the sums gives 1, 4, 9, 16, …, which are 1 squared, 2 squared, 3 squared, 4 squared; the chapter shows that each square number is the sum of the odd numbers up to that position, pictured as an L-shaped layer of dots wrapped around the previous square.
Why not A: the odd numbers are the numbers being added, not the running totals.
Why not C: triangular numbers 1, 3, 6, 10 come from adding consecutive natural numbers 1 + 2 + 3 + …, not consecutive odd numbers.
Why not D: the primes 2, 3, 5, 7, … follow no such addition pattern, and 4 and 9 are not prime.
Remember: naturals add up to triangles, odds add up to squares.
Why D: Step 1 (solve for 97): 5n – 3 = 97 gives 5n = 100, n = 20, a natural number, so 97 is the 20th term. Step 2 (solve for 172): 5n – 3 = 172 gives 5n = 175, n = 35, so 172 is the 35th term. Both numbers belong.
Why not A: both equations give natural-number positions, so both numbers are terms.
Why not B: 172 is also a term, since 172 + 3 = 175 = 5 × 35 gives the natural number n = 35.
Why not C: 97 is also a term, since 97 + 3 = 100 = 5 × 20 gives the natural number n = 20.
Remember: add the constant back, divide by the coefficient of n, and check that n is a natural number.
Why C: Step 1 (apply the rule repeatedly from t₁ = 2): t₂ = 3 × 2 – 2 = 4, t₃ = 3 × 4 – 2 = 10, t₄ = 3 × 10 – 2 = 28, t₅ = 3 × 28 – 2 = 82. Step 2 (continue until 730 appears): t₆ = 3 × 82 – 2 = 244, t₇ = 3 × 244 – 2 = 730. So 730 is the 7th term.
Why not A: 244 is the 6th term; stopping there miscounts by one.
Why not B: t₈ = 3 × 730 – 2 = 2188, far past 730.
Why not D: t₉ = 6562; counting the rule t_(n+1) as producing t_(n+2) leads to this overcount.
Remember: with t_(n+1) = …, n = 1 gives t₂; label each term as you compute it.
Why D: Step 1 (write both facts with the nth term formula): a + 2d = 12 and, since the last term is the 50th, a + 49d = 106. Step 2 (subtract the equations): 47d = 94, so d = 2 and a = 12 – 4 = 8. Step 3 (29th term): t₂₉ = 8 + 28 × 2 = 64.
Why not A: 60 = 8 + 26 × 2 adds d two times too few.
Why not B: 62 = 8 + 27 × 2 adds d only 27 times instead of 28.
Why not C: 66 = 8 + 29 × 2 uses n instead of n – 1 in the nth term.
Remember: two known terms give two linear equations in a and d; subtract to find d.
Why B: Step 1 (estimate): n(n + 1)/2 > 1000 means n(n + 1) > 2000, and 44 × 45 = 1980 while 45 × 46 = 2070, so n is around 45. Step 2 (check the exact sums): S₄₄ = 44 × 45/2 = 990, which is not greater than 1000; S₄₅ = 45 × 46/2 = 1035, which is. So the smallest such n is 45.
Why not A: S₄₄ = 990 is still below 1000, so 44 is one term too early.
Why not C: S₄₆ = 1081 does exceed 1000, but 45 already does, so 46 is not the smallest.
Why not D: 32 comes from solving n² > 1000 as if the sum were n².
Remember: estimate n from n(n + 1) > 2 × target, then verify the two neighbouring sums.
Why C: Step 1 (count how many times the ratio is applied): from the 8th term to the 12th term the ratio is multiplied 12 – 8 = 4 times, so t₁₂ = t₈ × 2⁴. Step 2 (arithmetic): 192 × 16 = 3072. Check via a: a = 192 ÷ 2⁷ = 1.5, and t₁₂ = 1.5 × 2¹¹ = 3072.
Why not A: 1536 = 192 × 2³ applies the ratio only 3 times, giving the 11th term.
Why not B: 6144 = 192 × 2⁵ applies it 5 times, giving the 13th term.
Why not D: 384 = 192 × 2 is just the 9th term.
Remember: tₘ = tₖ × r^(m – k); the exponent is the gap between the positions.
Why D: the list 6, 12, 24, 48, 96 stops after the fifth term; there are no three dots, so nothing follows 96, and the chapter gives exactly this list as an example of a finite sequence of five terms.
Why not A: the three dots after 6 mean the natural numbers continue indefinitely, so this is infinite.
Why not B: the odd numbers written with three dots continue forever.
Why not C: the square numbers with three dots also go on forever; writing only six of them does not make the sequence finite.
Remember: three dots at the end = infinite; a list that simply stops = finite.
Why B: Step 1 (set the rule equal to the number): 5n – 2 = 471 gives 5n = 473, so n = 94.6. Step 2 (position must be a natural number): 94.6 is not a natural number, there is no 94.6th term, so 471 is not a term of the sequence.
Why not A: 308 gives 5n = 310, n = 62, a natural number, so 308 is the 62nd term.
Why not C: 98 gives 5n = 100, n = 20, so 98 is the 20th term.
Why not D: 473 gives 5n = 475, n = 95, so 473 is the 95th term; do not confuse it with the 473 that appeared while testing 471.
Remember: solve for n; a term exists only when n is a natural number.
Why D: Step 1 (compute T₄ and T₅): T₄ = 4 + 2 + 1 = 7 and T₅ = 7 + 4 + 2 = 13. Step 2 (compute T₆ from the three terms before it): T₆ = 13 + 7 + 4 = 24. The sequence runs 1, 2, 4, 7, 13, 24, 44, 81.
Why not A: 13 is T₅, one step short.
Why not B: 20 = 13 + 7 adds only the previous two terms, but this rule adds three.
Why not C: 44 = 24 + 13 + 7 is T₇, one step too far.
Remember: read how many earlier terms the rule uses before you start adding.
Why B: Step 1 (identify a and d): a = 21, d = 18 – 21 = -3. Step 2 (set the nth term equal to -81 and solve): 21 + (n – 1)(-3) = -81 gives (n – 1)(-3) = -102, so n – 1 = 34 and n = 35. Since 35 is a natural number, -81 is the 35th term.
Why not A: 34 is the value of n – 1, not n; the count was stopped one step early.
Why not C: 36 adds 1 to the position n = 35 a second time; the 36th term is -84, not -81.
Why not D: the equation gives a natural number, so -81 is a term; negative values are perfectly possible in a decreasing AP.
Remember: solve for n – 1 first, then add 1 to get the position.
Chapter Navigation
Previous: Introduction to Probability Class 9 Notes
Next: Linear Equations in Two Variables Class 9 Notes
Related Chapters in Class 9 Maths
- Coordinate Geometry Class 9 Notes
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
- Exploring Algebraic Identities Class 9 Notes
- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
Explore More
Frequently Asked Questions
An explicit rule gives the nth term directly from its position n, for example tₙ = 2n − 1 for odd numbers, so you can find the 100th term in one step. A recursive rule gives each term from the previous term or terms, for example t₁ = 1, tₙ = tₙ₋₁ + 2 for n ≥ 2. A recursive rule must always state the starting term, otherwise it describes many different sequences.
Set the explicit rule equal to the number and solve for n. If n comes out as a natural number, the number is a term and n is its position. If n is a fraction, a decimal, zero or negative, the number is not a term. For example, 5n − 2 = 471 gives n = 94.6, so 471 is not a term of sₙ = 5n − 2.
The nth term of an arithmetic progression is tₙ = a + (n − 1)d, where a is the first term and d is the common difference. The first term already sits in position 1, so to reach position n you add d only n − 1 times. A quick check is to put n = 1: the formula must give back a.
Subtract each term from the next: if all the differences are equal, the sequence is an AP. Divide each term by the one before it: if all the ratios are equal, it is a GP. On a graph of position against term, AP points lie on a straight line while GP points lie on a curve that rises or falls faster and faster.
Write the sum forwards and backwards, one under the other. Each vertical pair, such as 1 + n and 2 + (n − 1), adds up to n + 1, and there are n pairs. So twice the sum is n(n + 1), and the sum is n(n + 1)/2. The first known written mention of this result is in Āryabhaṭa’s Āryabhaṭīya, Chapter 2, Verse 19.
Subtract two sums of natural numbers. For 25 + 26 + … + 58, take the sum up to 58 and remove the sum up to 24: S₅₈ − S₂₄ = 1711 − 300 = 1411. Subtract up to 24, not 25, so that 25 stays in the answer. Āryabhaṭa’s rule gives the same result: average of first and last terms times the number of terms.
At every stage each black triangle is replaced by three smaller ones, so the count is multiplied by 3: 1, 3, 9, 27, which is 3ⁿ at Stage n. Each stage also keeps only 3 of 4 equal parts, so the black area is multiplied by 3/4: 1, 3/4, 9/16, which is (3/4)ⁿ. The count grows quickly while the area shrinks towards 0.
