Class 9 Maths Ganita Manjari Chapter 8 Solutions (NCERT 2026–27) – Predicting What Comes Next: Exploring Sequences and Progressions

These Class 9 Maths Ganita Manjari Chapter 8 solutions cover Predicting What Comes Next: Exploring Sequences and Progressions from the new NCF-2023 textbook (2026–27). Every exercise is solved step by step — sequences, arithmetic progressions (AP) and geometric progressions (GP) — so you can revise the whole chapter quickly.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 8 Exercises: 8.1–8.3, End-of-Chapter Session: 2026–27
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Chapter 8 Overview

Chapter 8 of Ganita Manjari, Predicting What Comes Next: Exploring Sequences and Progressions, studies number patterns. It defines a sequence and its nth-term rule (explicit and recursive), then two special families: an arithmetic progression (AP) with constant common difference d and nth term a + (n − 1)d, and a geometric progression (GP) with constant common ratio r and nth term arn−1. The Class 9 Maths Ganita Manjari Chapter 8 solutions below work through every exercise step by step.

Key Concepts

Sequence: an ordered list of numbers; the nth term tn may be given by an explicit rule or a recursive rule.

Arithmetic progression (AP): consecutive terms differ by a constant common difference d.

Geometric progression (GP): consecutive terms have a constant common ratio r.

Explicit rule gives tn directly in terms of n; a recursive rule gives each term from the previous one(s).

Key Formulas

AP: a, a + d, a + 2d, … ; nth term tn = a + (n − 1)d.

GP: a, ar, ar2, … ; nth term tn = arn−1.

Common difference d = tn − tn−1; common ratio r = tn / tn−1.

Sum of the first n natural numbers = n(n + 1)/2.

Exercise Set 8.1

1. Find the first five terms of the sequence whose nth term is (i) tn = 3n − 4, (ii) tn = 2 − 5n, (iii) tn = n2 − 2n + 3 (n ≥ 1).

SOLUTION (i) −1, 2, 5, 8, 11. (ii) −3, −8, −13, −18, −23. (iii) 2, 3, 6, 11, 18.

2. Find the 10th and 15th terms of the sequence tn = 5n − 3 (n ≥ 1).

SOLUTIONt10 = 5(10) − 3 = 47; t15 = 5(15) − 3 = 72.

3. Determine whether 97 and 172 are terms of tn = 5n − 3 (n ≥ 1).

SOLUTION 5n − 3 = 97 ⇒ n = 20 (whole number) → 97 is the 20th term. 5n − 3 = 172 ⇒ n = 35 (whole number) → 172 is the 35th term.

4. Which term of tn = 5n − 3 (n ≥ 1) is 607?

SOLUTION5n − 3 = 607 ⇒ 5n = 610 ⇒ n = 122 → the 122nd term.

5. A sequence has t1 = −5, tn+1 = tn + 3 (n ≥ 1). Find the first five terms. Is 52 a term? If so, which?

SOLUTION First five: −5, −2, 1, 4, 7 (an AP with a = −5, d = 3, so tn = 3n − 8). 3n − 8 = 52 ⇒ n = 20 → yes, 52 is the 20th term.

6. Let T1 = 1, T2 = 2, T3 = 4, and Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 4. Find T4, T5, T6, T7, T8.

SOLUTION T4 = 4 + 2 + 1 = 7; T5 = 7 + 4 + 2 = 13; T6 = 13 + 7 + 4 = 24; T7 = 24 + 13 + 7 = 44; T8 = 44 + 24 + 13 = 81.

Exercise Set 8.2

1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, …

SOLUTIONa = 3, d = 5. t10 = 3 + 9(5) = 48; t26 = 3 + 25(5) = 128.

2. Which term of the AP 21, 18, 15, … is −81? Is 0 a term? Give reasons.

SOLUTION a = 21, d = −3, so tn = 24 − 3n. 24 − 3n = −81 ⇒ n = 35 → −81 is the 35th term. 24 − 3n = 0 ⇒ n = 8 (a whole number), so yes, 0 is the 8th term.

3. Find the nth term of the AP 11, 8, 5, 2, … and write its recursive rule.

SOLUTION a = 11, d = −3 ⇒ tn = 11 + (n − 1)(−3) = 14 − 3n. Recursive: t1 = 11, tn = tn−1 − 3.

4. An AP of 50 terms has 3rd term 12 and last term 106. Find the 29th term.

SOLUTION a + 2d = 12, a + 49d = 106 ⇒ 47d = 94 ⇒ d = 2, a = 8. t29 = a + 28d = 8 + 56 = 64.

5. How many 2-digit numbers are divisible by 3? What is their sum?

SOLUTION They are 12, 15, …, 99 (AP, a = 12, d = 3, last = 99): count = (99 − 12)/3 + 1 = 30. Sum = (n/2)(first + last) = (30/2)(12 + 99) = 15 × 111 = 1665.

6. Harish started at an annual salary of ₹5,00,000 with an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

SOLUTION 5,00,000 + 20,000k = 7,00,000 ⇒ 20,000k = 2,00,000 ⇒ k = 10 years (in the 11th year of work).

7. Marbles are arranged with 1 in the first row, 2 in the second, …, up to 25 rows. How many marbles in all?

SOLUTION 1 + 2 + … + 25 = 25 × 26 / 2 = 325 marbles.

Exercise Set 8.3

1. Find the 12th term of a GP with common ratio 2 whose 8th term is 192.

SOLUTION t8 = a·27 = 192 ⇒ a = 192/128 = 1.5. t12 = a·211 = 1.5 × 2048 = 3072.

2. Find the 10th and nth terms of the GP: 5, 25, 125, …

SOLUTION a = 5, r = 5 ⇒ tn = 5 × 5n−1 = 5n; t10 = 510 = 9,765,625.

*3. A sequence has t1 = 2, tn+1 = 3tn − 2 (n ≥ 1). Which term is 730?

SOLUTION Terms: 2, 4, 10, 28, 82, 244, 730, … (the rule gives tn = 3n−1 + 1). So 730 = 36 + 1 is the 7th term.

4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit and recursive formulas.

SOLUTION a = 2, r = 3 ⇒ 2·3n−1 = 4374 ⇒ 3n−1 = 2187 = 37 ⇒ n = 8th term. Explicit: tn = 2·3n−1; recursive: t1 = 2, tn = 3tn−1.

5. A ball dropped from 80 m bounces to 60% of its previous height each time. (i) What height does it reach after the 5th bounce? (ii) Total vertical distance travelled by the time it hits the ground the 6th time?

SOLUTION (i) Height after 5th bounce = 80 × (0.6)5 = 6.22 m (approx). (ii) Distance = first drop + 2 × (rise heights of bounces 1–5) = 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208) = 301.34 m (approx).

6. Which term of the sequence 2, 2√2, 4, … is 128?

SOLUTION a = 2, r = √2 ⇒ tn = 2·(√2)n−1. 128 = 2·(√2)n−1 ⇒ (√2)n−1 = 64 = (√2)12 ⇒ n = 13th term.

7. Sierpiński square carpet (Fig. 8.12): at each stage each shaded square is divided into 9 and the centre removed. (i) Red squares in Stages 0–3. (ii) Predict Stages 4 and 5. (iii) Rule for the nth stage (explicit + recursive). (iv) If Stage 0 area = 1, find the red area at Stages 1–5 and the nth stage; what happens as n increases?

SOLUTION (i) Red squares: Stage 0 = 1, Stage 1 = 8, Stage 2 = 64, Stage 3 = 512. (ii) Stage 4 = 84 = 4096; Stage 5 = 85 = 32768. (iii) Explicit: count = 8n; recursive: a0 = 1, an = 8an−1. (iv) Each stage keeps 8/9 of the area: areas are (8/9), (8/9)2, (8/9)3, (8/9)4, (8/9)5. Explicit (8/9)n; recursive A0 = 1, An = (8/9)An−1. As n increases, the area → 0.

Class 9 Maths Ganita Manjari Chapter 8 Solutions — End-of-Chapter Exercises

1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

SOLUTION a + 10d = 38, a + 15d = 73 ⇒ 5d = 35 ⇒ d = 7, a = −32. t31 = a + 30d = −32 + 210 = 178.

2. Determine the AP whose 3rd term is 16 and whose 7th term exceeds the 5th term by 12.

SOLUTION t7 − t5 = 2d = 12 ⇒ d = 6; a + 2d = 16 ⇒ a = 4. AP: 4, 10, 16, 22, …

*3. How many three-digit numbers are divisible by 7?

SOLUTION Smallest = 105, largest = 994 (AP, d = 7): count = (994 − 105)/7 + 1 = 128.

*4. How many multiples of 4 lie between 10 and 250?

SOLUTION First = 12, last = 248 (AP, d = 4): count = (248 − 12)/4 + 1 = 60.

*5. Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.

SOLUTION t5 = 4t3 ⇒ ar4 = 4ar2 ⇒ r2 = 4 ⇒ r = ±2. With a(1 + r) = −4: r = −2 ⇒ a = 4: GP 4, −8, 16, −32, … ; r = 2 ⇒ a = −4/3: GP −4/3, −8/3, −16/3, …

*6. Find all ways of expressing 100 as a sum of consecutive natural numbers.

SOLUTION 100 = 18 + 19 + 20 + 21 + 22 (5 terms) and 100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 (8 terms). (Apart from the trivial single term 100.)

*7. Bacteria double every hour, starting with 30. How many at the end of the 2nd hour, 4th hour and nth hour?

SOLUTION 2nd hour = 30·22 = 120; 4th hour = 30·24 = 480; nth hour = 30·2n.

*8. 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.

SOLUTION 2a + 10d = 24 and 2a + 14d = 44 ⇒ 4d = 20 ⇒ d = 5, a = −13. First three terms: −13, −8, −3.

*9. Find the smallest n such that the sum of the first n natural numbers exceeds 1000.

SOLUTION n(n + 1)/2 > 1000 ⇒ n(n + 1) > 2000. n = 44 gives 1980, n = 45 gives 2070, so the smallest n is 45.

*10. Which term of the GP: 2, 8, 32, … is 131072? Write the explicit and recursive formulas.

SOLUTION a = 2, r = 4 ⇒ 2·4n−1 = 131072 ⇒ 4n−1 = 65536 = 48 ⇒ n = 9th term. Explicit: tn = 2·4n−1; recursive: t1 = 2, tn = 4tn−1.

*11. The sum of the first three terms of a GP is 13/12 and their product is −1. Find the common ratio and the terms.

SOLUTION Take the terms a/r, a, ar. Product = a3 = −1 ⇒ a = −1. Sum: −(1/r + 1 + r) = 13/12 ⇒ r + 1/r = −25/12. 12r2 + 25r + 12 = 0 ⇒ r = −3/4 or −4/3. The terms are 4/3, −1, 3/4 (common ratio −3/4, or the reverse order with −4/3).

*12. If the 4th, 10th and 16th terms of a GP are x, y and z, prove that x, y, z are in GP.

SOLUTION x = ar3, y = ar9, z = ar15. Then y2 = a2r18 = (ar3)(ar15) = xz. Since y2 = xz, x, y, z are in GP. ✓

*13. The sum of the first three terms of a GP is 26 and the sum of their squares is 364. Find the terms.

SOLUTION a(1 + r + r2) = 26 and a2(1 + r2 + r4) = 364. Using 1 + r2 + r4 = (1 + r + r2)(1 − r + r2) gives (1 − r + r2)/(1 + r + r2) = 7/13 ⇒ 3r2 − 10r + 3 = 0 ⇒ r = 3 or 1/3. For r = 3, a = 2 ⇒ terms 2, 6, 18 (and r = 1/3 gives 18, 6, 2).

*14. P1 = 1, P2 = 2, and for n > 2, Pn = P1 + P2 + … + Pn−1 + 1. Find P1…P8; a simpler recursive and explicit formula.

SOLUTION 1, 2, 4, 8, 16, 32, 64, 128. Each term (from the 3rd) doubles the previous one: recursive Pn = 2Pn−1 (n ≥ 3); explicit Pn = 2n−1.

*15. W1 = 1, W2 = 2, and for n > 2, Wn = W1 + W2 + … + Wn−2 + 2. Find W1…W8. Do you recognise it?

SOLUTION 1, 2, 3, 5, 8, 13, 21, 34 — each term is the sum of the two before it. This is the Fibonacci (Virāhānka) sequence.

Common Mistakes to Avoid

Watch out for these

  • Using n instead of (n − 1) in tn = a + (n − 1)d or arn−1.
  • Confusing common difference (AP, subtract) with common ratio (GP, divide).
  • Sign errors with a negative common difference or ratio.
  • When checking if a value is a term, accept it only if n is a positive whole number.
  • For counting terms in an AP, use (last − first)/d + 1 — don’t forget the “+ 1”.
  • Mixing up explicit and recursive rules.

Practice MCQs & Assertion–Reason

1. The nth term of an AP with first term a and common difference d is:

(a) a + nd    (b) a + (n − 1)d    (c) arn−1    (d) a + (n + 1)d

2. The nth term of a GP with first term a and common ratio r is:

(a) a + (n − 1)r    (b) arn    (c) arn−1    (d) nar

3. The common difference of the AP 3, 8, 13, 18, … is:

(a) 3    (b) 5    (c) 8    (d) 2

4. The common ratio of the GP 2, 6, 18, … is:

(a) 2    (b) 3    (c) 4    (d) 6

5. The 10th term of the AP 2, 5, 8, … is:

(a) 27    (b) 29    (c) 30    (d) 32

6. Which sequence is an AP?

(a) 2, 4, 8, 16    (b) 1, 4, 9, 16    (c) 2, 4, 6, 8    (d) 1, 2, 4, 7

7. The 5th term of the GP 1, 2, 4, … is:

(a) 8    (b) 16    (c) 32    (d) 10

8. The 3rd term of an AP with a = 5 and d = −2 is:

(a) 1    (b) 3    (c) −1    (d) 7

9. The sum 1 + 2 + 3 + … + n equals:

(a) n2    (b) n(n + 1)    (c) n(n + 1)/2    (d) n/2

10. In a GP, the common ratio r equals:

(a) tn − tn−1    (b) tn + tn−1    (c) tn / tn−1    (d) tn−1 / tn

Answer key: 1-(b), 2-(c), 3-(b), 4-(b), 5-(b), 6-(c), 7-(b), 8-(a), 9-(c), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: 2, 4, 6, 8, … is an arithmetic progression.

Reason: The difference between consecutive terms is constant.

A-R 2. Assertion: 3, 6, 12, 24, … is a geometric progression.

Reason: The ratio of consecutive terms is constant.

A-R 3. Assertion: The nth term of an AP is a + (n − 1)d.

Reason: Each term is obtained by adding d to the previous term.

A-R 4. Assertion: 1, 2, 4, 7, 11, … is an arithmetic progression.

Reason: An arithmetic progression has a constant common difference.

A-R 5. Assertion: The common ratio of a GP can be negative.

Reason: A GP has the form a, ar, ar2, … where r may be any non-zero number.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • A sequence is an ordered list with an nth-term rule (explicit or recursive).
  • AP: constant common difference d; tn = a + (n − 1)d.
  • GP: constant common ratio r; tn = arn−1.
  • To check if a value is a term, solve for n and require a positive whole number.
  • Number of terms of an AP = (last − first)/d + 1.
  • The Fibonacci (Virāhānka) sequence adds the two previous terms: 1, 2, 3, 5, 8, …

How to score full marks in this chapter

Identify a and d (for AP) or a and r (for GP) first, then apply the nth-term formula. Write both the explicit and the recursive rule when asked. For “which term” problems, set tn equal to the value and check that n is a positive integer.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 8 about?

Sequences and progressions — the nth-term rule of a sequence, arithmetic progressions (AP) and geometric progressions (GP).

What is the nth term of an AP?

tn = a + (n − 1)d, where a is the first term and d the common difference.

What is the nth term of a GP?

tn = arn−1, where a is the first term and r the common ratio.

Are these Class 9 Maths Ganita Manjari Chapter 8 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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