1. Chapter at a Glance
2, 5, 8, 11, … what comes next? 3, 6, 12, 24, …? 1, 1, 2, 3, 5, 8, …? Spotting the rule behind a list of numbers lets us predict any later term without writing them all out. In this chapter we study sequences — ordered lists of numbers — and two very important kinds: arithmetic progressions (add the same number each time) and geometric progressions (multiply by the same number each time). We also meet recursive rules such as the Virahāṅka–Fibonacci sequence, and connect them to visual patterns and fractals.
- Sequences, terms and the general (nth) term.
- Explicit rules vs recursive rules.
- Arithmetic progressions (AP): common difference, nth term, sum of n terms.
- Geometric progressions (GP): common ratio, nth term, sum of n terms.
- The Virahāṅka–Fibonacci sequence.
- Growth: linear (AP) vs exponential (GP).
2. Sequences and Rules
A sequence is a list of numbers in a definite order: a₁, a₂, a₃, …, where aₙ is the nth term.
- Explicit rule — gives aₙ directly from n. Example: aₙ = n² gives 1, 4, 9, 16, …
- Recursive rule — gives each term from earlier terms. Example: a₁ = 3, aₙ = aₙ₋₁ + 4 gives 3, 7, 11, 15, …
Write the first four terms of aₙ = 2n² − 1, and of a₁ = 2, aₙ = 3aₙ₋₁.
Solution: 1, 7, 17, 31. 2, 6, 18, 54.
3. Arithmetic Progressions
An AP is a sequence where each term is obtained by adding a fixed number d, the common difference, to the previous term: a, a + d, a + 2d, a + 3d, …
aₙ = a + (n − 1)d
To reach the nth term from the 1st, we take (n − 1) steps of size d.
Find the 30th term of 7, 11, 15, 19, …
Solution: a = 7, d = 4. a₃₀ = 7 + 29 × 4 = 123.
Which term of 3, 8, 13, … is 248?
Solution: 3 + (n − 1)5 = 248 → n − 1 = 49 → n = 50.
The 4th term of an AP is 13 and the 10th term is 31. Find a and d.
Solution: a + 3d = 13, a + 9d = 31. Subtract: 6d = 18 → d = 3, a = 4. AP: 4, 7, 10, …
The terms of an AP plotted against n lie on a straight line — an AP is a linear pattern (aₙ = dn + (a − d)).
4. Sum of an AP
A famous story says the young Gauss added 1 + 2 + … + 100 instantly by pairing: (1 + 100) + (2 + 99) + … = 50 pairs of 101 = 5050. The same pairing works for any AP.
Sₙ = n/2 × [2a + (n − 1)d] = n/2 × (first term + last term)
Find the sum of the first 20 terms of 5, 9, 13, …
Solution: S₂₀ = 10 × [10 + 19 × 4] = 10 × 86 = 860.
Meena saves ₹50 in week 1, ₹60 in week 2, ₹70 in week 3, and so on. How much does she save in 12 weeks?
Solution: S₁₂ = 6 × [100 + 11 × 10] = 6 × 210 = ₹1,260.
Useful: sum of the first n odd numbers 1 + 3 + 5 + … = n².
5. Geometric Progressions
A GP is a sequence where each term is obtained by multiplying the previous one by a fixed non-zero number r, the common ratio: a, ar, ar², ar³, …
aₙ = arⁿ⁻¹
Sₙ = a(rⁿ − 1)/(r − 1) for r ≠ 1 (equivalently a(1 − rⁿ)/(1 − r)).
Find the 8th term and the sum of the first 8 terms of 3, 6, 12, …
Solution: a = 3, r = 2. a₈ = 3 × 2⁷ = 384. S₈ = 3(2⁸ − 1)/1 = 765.
A ball dropped from 64 m bounces back to half its height each time. Find the height after the 5th bounce.
Solution: Heights after bounces: 32, 16, 8, 4, 2 → 2 m (a = 32, r = ½, a₅ = 32 × (½)⁴).
1 grain on the first square, 2 on the second, 4 on the third, … doubling each time. The 64th square alone gets 2⁶³ grains, and the total is 2⁶⁴ − 1 ≈ 1.8 × 10¹⁹ — more rice than the whole world produces. GP growth is enormous!
6. Linear vs Exponential Growth
| n | 1 | 2 | 3 | 4 | 5 | 6 | 8 | 10 |
|---|---|---|---|---|---|---|---|---|
| AP: 10, 20, 30, … | 10 | 20 | 30 | 40 | 50 | 60 | 80 | 100 |
| GP: 1, 2, 4, … | 1 | 2 | 4 | 8 | 16 | 32 | 128 | 512 |
The GP starts far behind but overtakes the AP by n = 8 and then races away.
7. The Virahāṅka–Fibonacci Sequence
1, 1, 2, 3, 5, 8, 13, 21, 34, … Each term is the sum of the two before it: Fₙ = Fₙ₋₁ + Fₙ₋₂.
The Indian scholar Virahāṅka (around 600–800 CE) found this sequence while counting the rhythms of Sanskrit poetry that can be made from short (1 beat) and long (2 beat) syllables. The number of rhythms with n beats follows exactly this rule — centuries before Fibonacci (1202 CE) used it for rabbit populations. It appears in sunflower seed spirals, pine cones and pineapple scales.
Ratios of consecutive terms (8/5, 13/8, 21/13, …) get closer and closer to the golden ratio ≈ 1.618.
8. Solved Problems (Board Pattern)
How many two-digit numbers are divisible by 7? Find their sum.
Solution: 14, 21, …, 98: 14 + (n − 1)7 = 98 → n = 13. Sum = 13/2 × (14 + 98) = 728.
An auditorium has 20 seats in the first row, and each row has 2 more seats than the one in front. How many seats are there in 25 rows?
Solution: S₂₅ = 25/2 × [40 + 24 × 2] = 25/2 × 88 = 1100.
The 3rd term of a GP is 18 and the 6th term is 486. Find a and r.
Solution: ar⁵/ar² = r³ = 27 → r = 3; a = 18/9 = 2.
9. Master Formula Table
| Result | Formula |
|---|---|
| AP common difference | d = aₙ₊₁ − aₙ |
| AP nth term | aₙ = a + (n − 1)d |
| AP sum | Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l) |
| 1 + 2 + … + n | n(n + 1)/2 |
| Sum of first n odd numbers | n² |
| GP common ratio | r = aₙ₊₁ / aₙ |
| GP nth term | aₙ = arⁿ⁻¹ |
| GP sum | Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1 |
| Fibonacci rule | Fₙ = Fₙ₋₁ + Fₙ₋₂ |
10. Master Summary Table
| Feature | AP | GP |
|---|---|---|
| Step rule | Add d | Multiply by r |
| Test | Differences constant | Ratios constant |
| nth term | a + (n − 1)d | arⁿ⁻¹ |
| Graph against n | Straight line | Curve (fast growth if r > 1) |
| Real life | Savings, seats, taxi fares | Doubling, halving, compound growth |
Chart — 10th terms of three sequences:
11. Mind Map
- Explicit
- Recursive
- a + (n − 1)d
- n/2 (a + l)
- arⁿ⁻¹
- a(rⁿ − 1)/(r − 1)
- Virahāṅka–Fibonacci
- Golden ratio
12. Tips and Tricks
- Check differences first; if not constant, check ratios.
- When two AP terms are given, subtract the equations to find d quickly.
- For the sum of an AP when the last term is known, use n/2 (first + last).
- Using n instead of (n − 1) in aₙ = a + (n − 1)d.
- Finding d as a₁ − a₂ instead of a₂ − a₁ (sign error).
- Counting terms: from 14 to 98 in steps of 7 there are 13 terms, not 12.
13. Important One-liners
- An AP grows by adding; a GP grows by multiplying.
- The nth term of an AP is a linear expression in n.
- 1 + 2 + … + n = n(n + 1)/2.
- Sum of the first n odd numbers is n².
- Virahāṅka described the "Fibonacci" sequence centuries before Fibonacci.
- A GP with r > 1 eventually overtakes any AP.
14. Exam-pattern Questions with Detailed Answers
Find the common difference of 15, 11, 7, 3, …
Show answer
11 − 15 = −4.
Answer: −4
Which of these is a GP?
- 2, 4, 6, 8
- 5, 15, 45, 135
- 1, 4, 9, 16
- 3, 5, 8, 13
Show answer
Ratio 3 throughout in (B).
Answer: (B)
Find the 15th term of 2, 7, 12, …
Show answer
2 + 14 × 5 = 72.
Answer: 72
Find the sum of the first 50 natural numbers and the sum of the first 50 odd numbers.
Show answer
50 × 51/2 = 1275; 50² = 2500.
Answer: 1275 and 2500
Is 301 a term of the AP 5, 11, 17, …? Justify.
Show answer
5 + (n − 1)6 = 301 → n − 1 = 296/6 = 49.33…, not a whole number. So 301 is not a term.
Answer: No
Find the 7th term of the GP 81, 27, 9, …
Show answer
a = 81, r = 1/3. a₇ = 81 × (1/3)⁶ = 81/729 = 1/9.
Answer: 1/9
Write the next three terms: 1, 3, 4, 7, 11, …, and state the rule.
Show answer
Each term = sum of the previous two: 18, 29, 47.
Answer: 18, 29, 47
The 6th term of an AP is 19 and the 15th term is 46. Find the 25th term.
Show answer
9d = 27 → d = 3; a = 19 − 15 = 4. a₂₅ = 4 + 72 = 76.
Answer: 76
How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?
Show answer
n/2 [18 + 8(n − 1)] = 636 → n(4n + 5) = 636 → 4n² + 5n − 636 = 0 → (n − 12)(4n + 53) = 0 → n = 12.
Answer: 12 terms
A bacterium splits into 2 every 20 minutes. Starting with one, how many bacteria are there after 4 hours?
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4 hours = 12 doublings. 2¹² = 4096.
Answer: 4096
Find the sum of all three-digit numbers that leave remainder 2 when divided by 5.
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102, 107, …, 997. n: 102 + (n − 1)5 = 997 → n = 180. Sum = 90 × (102 + 997) = 98910.
Answer: 98910
Count the rhythms of 5 beats made from short (1-beat) and long (2-beat) syllables, and explain why the counts follow the Virahāṅka–Fibonacci rule.
Show answer
A rhythm of n beats ends either in a short syllable (the rest is a rhythm of n − 1 beats) or a long one (the rest has n − 2 beats). So Rₙ = Rₙ₋₁ + Rₙ₋₂. With R₁ = 1 and R₂ = 2: R₃ = 3, R₄ = 5, R₅ = 8.
Answer: 8 rhythms
15. Quick Revision Summary
- Sequences: explicit (aₙ from n) or recursive (from earlier terms).
- AP: aₙ = a + (n − 1)d; Sₙ = n/2 [2a + (n − 1)d].
- GP: aₙ = arⁿ⁻¹; Sₙ = a(rⁿ − 1)/(r − 1).
- AP = linear growth; GP = exponential growth.
- Virahāṅka–Fibonacci: Fₙ = Fₙ₋₁ + Fₙ₋₂; ratios → golden ratio.