Class 9 Maths · Chapter 8
Short answer:
Class 9 Maths Chapter 8 — Predicting What Comes Next: Exploring Sequences and Progressions — Ganita Manjari ka bilkul naya chapter hai (session 2026-27 se), jisme number patterns ko formal language milti hai: general sequences (explicit aur recursive rules), Arithmetic Progression (AP), aur Geometric Progression (GP). Ye ganita manjari class 9 maths ncert solutions series ka last (8th) chapter hai aur seedha Class 10 ki AP unit ki neev banata hai.
Chapter 8 poora naya hai — pehle Class 9 me is naam ka koi chapter nahi hota tha; Arithmetic aur Geometric Progressions purane syllabus me sirf Class 10 me aati thi. NCF-2023-based naye textbook Ganita Manjari Part 1 ne ise Class 9 me hi laa diya hai, taaki students pattern-thinking early build karein. Chapter Class 6-7-8 ke informal number-pattern experience ko utha kar teen formal structures me daalta hai: (1) general sequences — jinka rule explicit (direct formula) ya recursive (pichle term se agla term) ho sakta hai, jisme historic Virahānka–Fibonacci sequence bhi cover hoti hai; (2) Arithmetic Progression (AP) — fixed common difference wali sequence, jaise salary increment ya stadium seating; (3) Geometric Progression (GP) — fixed common ratio wali sequence, jaise bouncing ball ki height ya bacteria growth, aur Sierpiński-jaisi fractal patterns.
Yeh chapter class 9 maths new syllabus 2026-27 chapter list pdf me Ganita Manjari Part 1 ka aakhri (8th) chapter hai — is Part 1 me total 8 chapters hi published hain, Part 2 abhi tak release nahi hua. Neeche in-text practice (Exercise 8.1 style — general sequences) aur main exercises (8.2 AP + 8.3 GP style) dono full step-by-step working ke saath diye gaye hain, taaki poora method clearly dikhe, sirf final answer nahi.
Chapter 8 Summary — 5 Minute Revision
Chapter 8 ka core takeaway simple hai: har pattern ke peeche ek rule hota hai, aur woh rule ya to explicit hota hai (directly n se tn nikaalo) ya recursive (pichle term(s) se agla term nikaalo). AP me yeh rule fixed addition (common difference d) hai — tₙ = a + (n−1)d — aur GP me fixed multiplication (common ratio r) hai — tₙ = ar^(n−1). Sum formulas (Sₙ for AP aur GP) real-life applications (salary, seating, bouncing ball, bacteria growth) solve karne ke liye use hoti hain. Yehi neev seedha Class 10 Maths ki poori AP unit me carry hoti hai — jo student yahan explicit vs recursive aur AP vs GP ka farak clearly samajh leta hai, uske liye Class 10 me AP ek revision jaisa lagega, naya topic nahi.
In-Text Questions — Solutions
Neeche diye pattern ke agle 3 terms likhiye, aur bataiye ki yeh recursive rule follow karta hai ya explicit: 3, 7, 11, 15, ...
Step 1 — Pattern check:
7 − 3 = 4, 11 − 7 = 4, 15 − 11 = 4
Har baar difference constant (4) hai, isliye recursive rule likha ja sakta hai:
t₁ = 3, tₙ = tₙ₋₁ + 4 (n > 1)
Step 2 — Agle 3 terms:
t₅ = 15 + 4 = 19, t₆ = 19 + 4 = 23, t₇ = 23 + 4 = 27
Step 3 — Explicit rule (verify):
tₙ = 3 + 4(n − 1) = 4n − 1 ⟹ t₅ = 4(5) − 1 = 19 ✓
Answer: agle terms 19, 23, 27; explicit rule tₙ = 4n − 1.
Ek sequence recursive rule se di gayi hai: t₁ = 1, t₂ = 1, tₙ = tₙ₋₁ + tₙ₋₂ (n ≥ 3). Iske pehle 8 terms likhiye. Ye kaunsi famous sequence hai?
Step-by-step terms:
t₁ = 1, t₂ = 1
t₃ = t₂ + t₁ = 1 + 1 = 2
t₄ = t₃ + t₂ = 2 + 1 = 3
t₅ = t₄ + t₃ = 3 + 2 = 5
t₆ = t₅ + t₄ = 5 + 3 = 8
t₇ = t₆ + t₅ = 8 + 5 = 13
t₈ = t₇ + t₆ = 13 + 8 = 21
Sequence: 1, 1, 2, 3, 5, 8, 13, 21. Yeh Virahānka–Fibonacci sequence hai — Acharya Virahānka ne isse Fibonacci se sadiyon pehle Sanskrit prosody (chhand-shastra) me describe kiya tha.
Sequence 2, 4, 8, 16, 32, ... ka explicit rule aur recursive rule dono likhiye.
Pattern check: Har term pichle term ka double hai (4÷2=2, 8÷4=2, 16÷8=2 — constant ratio 2).
Recursive rule:
t₁ = 2, tₙ = 2 × tₙ₋₁ (n > 1)
Explicit rule:
tₙ = 2ⁿ
Verify: t₁ = 2¹ = 2 ✓, t₅ = 2⁵ = 32 ✓
Ek sequence recursive rule se di gayi hai: t₁ = 5, tₙ = tₙ₋₁ + 3 (n > 1). Explicit rule likhiye aur t₁₀ nikaliye.
Step 1 — Constant difference: Har baar 3 add ho raha hai, isliye:
tₙ = 5 + 3(n − 1) = 3n + 2
Step 2 — Verify with recursion: t₂ = t₁ + 3 = 5 + 3 = 8; explicit se t₂ = 3(2)+2 = 8 ✓
Step 3 — t₁₀:
t₁₀ = 3(10) + 2 = 32
Answer: explicit rule tₙ = 3n + 2, t₁₀ = 32.
Triangular numbers ka pattern hai 1, 3, 6, 10, 15, ... Iska explicit rule likhiye aur 10th triangular number nikaliye.
Step 1 — Differences dekhiye: 3−1=2, 6−3=3, 10−6=4, 15−10=5 — difference khud badh raha hai (2,3,4,5,...), isliye yeh AP nahi hai, ek quadratic pattern hai.
Step 2 — Known rule (triangular numbers):
tₙ = n(n + 1) / 2
Step 3 — Verify: t₄ = 4(5)/2 = 10 ✓
Step 4 — t₁₀:
t₁₀ = 10(11) / 2 = 110 / 2 = 55
Answer: tₙ = n(n+1)/2, 10th triangular number = 55.
Sequence ka explicit rule diya hai: tₙ = 2n − 1 (yeh odd numbers ki sequence hai). Iska recursive rule likhiye.
Step 1 — Pehla term:
t₁ = 2(1) − 1 = 1
Step 2 — Common difference: Consecutive odd numbers ka difference hamesha 2 hota hai.
Recursive rule:
t₁ = 1, tₙ = tₙ₋₁ + 2 (n > 1)
Verify: Explicit se t₂ = 2(2)−1 = 3; recursive se t₂ = t₁+2 = 1+2 = 3 ✓
Ek fractal design (Sierpiński triangle jaisa) me shaded triangles ka pattern hai: 1st iteration me 1, 2nd me 3, 3rd me 9, 4th me 27 shaded triangle. Pattern ka explicit rule likhiye aur 6th iteration me shaded triangles ki sankhya nikaliye.
Step 1 — Ratio check:
3÷1 = 3, 9÷3 = 3, 27÷9 = 3 — constant ratio 3
Step 2 — Explicit rule:
tₙ = 1 × 3^(n−1) = 3^(n−1)
Step 3 — 6th iteration (t₆):
t₆ = 3^(6−1) = 3⁵ = 243
Answer: 6th iteration me 243 shaded triangles honge.
Ek tiling border design me squares ka pattern hai: 4, 7, 10, 13, ... Iska explicit (nth term) rule likhiye aur 12th step me kitne squares honge, nikaliye.
Step 1 — Constant difference:
7−4 = 3, 10−7 = 3, 13−10 = 3 (d = 3)
Step 2 — Explicit rule:
tₙ = 4 + 3(n − 1) = 3n + 1
Step 3 — t₁₂:
t₁₂ = 3(12) + 1 = 36 + 1 = 37
Answer: 12th step me 37 squares honge.
Exercise Questions — Solutions (Q1–Q14)
Ek AP ka first term a = 5 aur common difference d = 3 hai. nth term ka formula likhiye aur 20th term (t₂₀) nikaliye.
Formula:
tₙ = a + (n − 1)d = 5 + (n − 1)(3) = 3n + 2
t₂₀:
t₂₀ = 3(20) + 2 = 60 + 2 = 62
Answer: tₙ = 3n + 2, t₂₀ = 62.
Ek AP me a₃ = 7 aur a₈ = 22 hai. Common difference d aur first term a nikaliye, phir AP likhiye.
Step 1 — Equations banao:
a + 2d = 7 ... (i)
a + 7d = 22 ... (ii)
Step 2 — (ii) − (i):
5d = 15 ⟹ d = 3
Step 3 — a nikaliye:
a + 2(3) = 7 ⟹ a = 1
Answer: a = 1, d = 3, AP = 1, 4, 7, 10, 13, ...
AP 3, 8, 13, 18, ... ka kaunsa term 88 hai?
Given: a = 3, d = 5, tₙ = 88
tₙ = a + (n − 1)d
88 = 3 + (n − 1)(5)
85 = 5(n − 1) ⟹ n − 1 = 17 ⟹ n = 18
Answer: 88, AP ka 18th term hai.
AP 2, 5, 8, 11, ... ke first 20 terms ka sum nikaliye.
Given: a = 2, d = 3, n = 20
Sₙ = n/2 [2a + (n − 1)d]
S₂₀ = 20/2 [2(2) + 19(3)] = 10[4 + 57] = 10 × 61
S₂₀ = 610
Answer: Sum = 610.
Harish ki yearly salary review AP follow karti hai — starting figure ₹20,000 hai aur har saal fixed ₹1,500 ka increment milta hai. (a) 8th year ki figure kya hogi? (b) Pehle 10 saal ki figures ka total (sum) kya hoga?
Given: a = 20000, d = 1500
(a) 8th year (t₈):
t₈ = a + 7d = 20000 + 7(1500) = 20000 + 10500 = 30500
(b) Sum of 10 terms (S₁₀):
S₁₀ = 10/2 [2(20000) + 9(1500)] = 5[40000 + 13500] = 5 × 53500
S₁₀ = 267500
Answer: 8th year ki figure = ₹30,500; pehle 10 saal ki figures ka sum = ₹2,67,500.
Ek stadium section ki pehli row me 20 seats hain aur har agli row me 4 seats zyada hain. Section me total 15 rows hain. Last (15th) row me kitni seats hain, aur poore section me total kitni seats hain?
Given: a = 20, d = 4, n = 15
Last row (t₁₅):
t₁₅ = 20 + 14(4) = 20 + 56 = 76
Total seats (S₁₅):
S₁₅ = 15/2 [2(20) + 14(4)] = 15/2 [40 + 56] = 15/2 × 96
S₁₅ = 15 × 48 = 720
Answer: last row me 76 seats; total 720 seats.
Teen numbers AP me hain. Unka sum 27 hai aur product 288 hai. Numbers nikaliye.
Step 1 — Numbers ko (a−d), a, (a+d) le lo:
Sum: (a−d) + a + (a+d) = 27 ⟹ 3a = 27 ⟹ a = 9
Step 2 — Product:
(9−d)(9)(9+d) = 288
9(81 − d²) = 288 ⟹ 81 − d² = 32 ⟹ d² = 49 ⟹ d = ±7
Step 3 — Numbers: d = 7 lene par (2, 9, 16); d = −7 lene par (16, 9, 2) — same set.
Answer: Numbers 2, 9, 16.
Ek GP ka first term a = 3 aur common ratio r = 2 hai. nth term ka formula likhiye aur 7th term (t₇) nikaliye.
Formula:
tₙ = ar^(n−1) = 3 × 2^(n−1)
t₇:
t₇ = 3 × 2⁶ = 3 × 64 = 192
Answer: tₙ = 3 × 2^(n−1), t₇ = 192.
Ek GP me a₂ = 6 aur a₅ = 48 hai. Common ratio r aur first term a nikaliye, phir first 5 terms likhiye.
Step 1 — Equations:
ar = 6 ... (i)
ar⁴ = 48 ... (ii)
Step 2 — (ii) ÷ (i):
r³ = 48/6 = 8 ⟹ r = 2
Step 3 — a nikaliye:
a(2) = 6 ⟹ a = 3
Answer: a = 3, r = 2, terms = 3, 6, 12, 24, 48
GP 2, 6, 18, 54, ... ke first 6 terms ka sum nikaliye.
Given: a = 2, r = 3, n = 6
Sₙ = a(rⁿ − 1)/(r − 1)
S₆ = 2(3⁶ − 1)/(3 − 1) = 2(729 − 1)/2 = 2(728)/2
S₆ = 728
Answer: Sum = 728.
Ek ball 8 m ki height se giraayi jaati hai aur har bounce par pichli height ka aadha (1/2) hi wapas uchhalti hai. 4th bounce ki height kya hogi?
Setup: Bounce heights GP banati hain jiska first term (1st bounce) a = 8 × 1/2 = 4, aur common ratio r = 1/2.
Height after nth bounce = 8 × (1/2)ⁿ
4th bounce:
t₄ = 8 × (1/2)⁴ = 8/16 = 0.5 m
Answer: 4th bounce ki height = 0.5 m.
Ek bacteria culture me shuru me 500 bacteria hain, aur har ghante me sankhya double ho jaati hai. 6 ghante baad kitne bacteria honge?
Setup: GP: a = 500, r = 2
Population after n hours = a × rⁿ = 500 × 2ⁿ
n = 6:
500 × 2⁶ = 500 × 64 = 32000
Answer: 6 ghante baad 32,000 bacteria honge.
GP 2, −6, 18, −54, ... ka 6th term (t₆) nikaliye.
Given: a = 2, r = −3 (negative ratio — sign alternate ho raha hai)
tₙ = ar^(n−1)
t₆ = 2 × (−3)⁵ = 2 × (−243) = −486
Answer: t₆ = −486 (negative — kyunki 5 (odd power) baar negative ratio multiply hua).
Ek fractal pattern me har iteration me shaded triangles pichli iteration se 3 guna ho jaate hain. Pehli iteration me 1 triangle shaded hai. 5th iteration me kitne triangles shaded honge, aur pehli 5 iterations ke total shaded triangles (sum) kitne honge?
Given: GP with a = 1, r = 3, n = 5
5th iteration (t₅):
t₅ = 1 × 3⁴ = 81
Sum of first 5 iterations (S₅):
S₅ = a(rⁿ − 1)/(r − 1) = 1(3⁵ − 1)/(3 − 1) = (243 − 1)/2 = 242/2
S₅ = 121
Answer: 5th iteration me 81 triangles; total 121 triangles.
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| General sequence — explicit rule | tₙ = f(n) — direct formula se koi bhi term nikal sakte ho (jaise tₙ = 2n − 1) |
| General sequence — recursive rule | t₁ (base term diya hota hai), tₙ = g(tₙ₋₁, tₙ₋₂, ...) — pichle term(s) se agla term |
| Virahānka–Fibonacci sequence | t₁ = 1, t₂ = 1, tₙ = tₙ₋₁ + tₙ₋₂ (n ≥ 3) |
| AP — nth term | tₙ = a + (n − 1)d, jahan a = first term, d = common difference |
| AP — common difference | d = t₂ − t₁ = t₃ − t₂ = ... (constant) |
| AP — sum of first n terms | Sₙ = n/2 [2a + (n − 1)d] = n/2 [a + l], jahan l = last term |
| Sum of first n natural numbers | 1 + 2 + 3 + ... + n = n(n + 1)/2 (triangular number formula) |
| GP — nth term | tₙ = a r^(n − 1), jahan a = first term, r = common ratio |
| GP — common ratio | r = t₂ / t₁ = t₃ / t₂ = ... (constant, r ≠ 0) |
| GP — sum of first n terms (r ≠ 1) | Sₙ = a(rⁿ − 1)/(r − 1) [r > 1] ya Sₙ = a(1 − rⁿ)/(1 − r) [r < 1] |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
AP aur GP ka farak bhool jaana: GP me consecutive terms ko subtract karke 'd' dhoondhna galat hai — GP me terms ko divide karke common ratio r nikalte hain, subtract karke nahi.
(n − 1) me off-by-one error: tₙ = a + (n − 1)d ki jagah tₙ = a + nd likh dena — result me har term ek 'd' zyada/kam aa jaata hai. Hamesha (n − 1) power/multiplier check karo.
Recursive rule ka base term likhna bhool jaana: sirf tₙ = tₙ₋₁ + d likhna adhoora hai — t₁ (ya jo bhi base term ho) explicitly likhna zaroori hai, warna rule define hi nahi hota.
Negative common ratio me sign error: GP me r negative ho (jaise r = −3) to odd power par answer negative aur even power par positive aata hai — students galti se hamesha positive answer likh dete hain.
AP ka sum formula GP problems me use kar dena (ya ulta): word problems me pehle check karo ki pattern additive hai (AP) ya multiplicative hai (GP), tabhi sahi Sₙ formula chuno.
Term-position confusion (t₁ se shuru ya t₀ se): 'nth bounce' ya 'n hours baad' jaise word problems me first term kya represent karta hai (starting value ya 1st change ke baad ka value) — ye clearly identify na karne se off-by-one answer aata hai.
Board-Style Important Questions
- Sequence 5, 8, 11, 14, ... ka explicit rule likhiye aur bataiye ki yeh AP hai ya GP.
- Ek GP me a = 4 aur r = 3 hai. t₅ nikaliye.
- Ek AP ka a₅ = 18 aur a₉ = 34 hai. Common difference aur first term nikaliye.
- AP 4, 9, 14, 19, ... ke first 15 terms ka sum nikaliye.
- Ek GP me a₃ = 12 aur a₆ = 96 hai. First term aur common ratio nikaliye, phir first 6 terms ka sum nikaliye.
Aksar Poochhe Jaane Wale Sawaal
Chapter 8 Sequences and Progressions Class 9 me bilkul naya kyun hai — pehle to yeh sirf Class 10 me hota tha?
Sahi observation hai. Purane NCERT syllabus me AP/GP sirf Class 10 me tha. NCF-2023 based Ganita Manjari ne pattern-thinking ko early introduce karne ke liye ise Class 9 me shift kar diya — yeh wahi number-pattern intuition ko formalize karta hai jo Class 9 Maths Chapter 3 The World of Numbers NCERT solutions aur Class 9 Maths Chapter 1 Orienting Yourself Use of Coordinates solutions jaise chapters me bhi background me use hoti hai.
Class 9 Maths Ganita Manjari Part 2 kab aayega, aur kya usme koi progression-related topic hoga?
As of 4 Aug 2026, Ganita Manjari Part 2 release nahi hua hai — official release date bhi announce nahi hui. Multiple secondary sources ke hisaab se Part 2 me expected topics Linear Equations in Two Variables, Euclid's Geometry, Lines and Angles, Triangles, Quadrilaterals, Surface Area and Volume, aur Statistics hain — koi naya progression-topic expected nahi hai. Jab tak official confirmation na aaye, is par content banana avoid karein.
Class 9 Maths ka pura naya syllabus 2026-27 chapter list kahan milega?
Class 9 Maths new syllabus 2026-27 chapter list pdf me Ganita Manjari Part 1 ke 8 chapters hain: Orienting Yourself, Introduction to Linear Polynomials, The World of Numbers, Exploring Algebraic Identities, I'm Up and Down and Round and Round (Circles), Measuring Space, Probability, aur Sequences and Progressions (ye chapter). Part 1 ke saare 8 chapters ka combined content class 9 maths ganita manjari all chapters pdf download ke through cover kiya ja sakta hai.
Sequences and Progressions ke sawaal exam me kis tarah mix ho sakte hain — kya isme algebra ka use hota hai?
Haan, khaaskar AP/GP ke word problems me linear equations solve karne padte hain (jaise a aur d dono nikalna). Ye skill Class 9 Maths Chapter 4 Exploring Algebraic Identities extra questions aur Class 9 Maths Chapter 2 Introduction to Linear Polynomials important questions me practice ki gayi algebra se directly connect hoti hai — dono chapters revise karna helpful rahega.
Kya Chapter 8 ka koi connection Chapter 6 aur Chapter 7 se bhi hai?
Direct syllabus overlap nahi hai, par exam-pattern me combined/case-study questions aa sakte hain jisme ek hi real-life scenario me progression aur measurement dono ho — isliye Class 9 Maths Chapter 6 Measuring Space Perimeter and Area important questions aur Class 9 Maths Chapter 7 Probability NCERT solutions dono ko parallel revise karna faydemand hai.
Fractal patterns (jaise Sierpiński triangle) ka Chapter 5 Circles se koi lena-dena hai kya?
Nahi, dono alag concepts hain — fractal yahan sirf GP ka ek visual/real-life example hai (har iteration me shaded shapes ek fixed ratio se badhte hain). Circles ka geometry topic Class 9 Maths Chapter 5 I'm Up and Down and Round and Round Circles important questions me alag se cover hota hai, iska progression se koi formula-level overlap nahi hai.
Class 9 Maths — Saare Chapters
NCERT Kaksha ek swatantra shaikshik platform hai aur NCERT ya CBSE se aadhikarik roop se sambaddh (officially affiliated) nahi hai. Kisi galti ki jaankari dene ke liye contact kijiye.