Exploring Sequences and Progressions for Class 9: Important Questions and Answers on Predicting What Comes Next

Important Questions on Predicting What Comes Next: Exploring Sequences and Progressions for Class 9 are provided in this Maths article to help students understand patterns, sequences, and logical number arrangements. These questions cover important topics such as identifying patterns, finding missing terms, arithmetic progressions (AP), geometric progressions (GP), and rule-based sequences with solved examples and exam-oriented explanations based on the CBSE syllabus and NCERT textbook. Whether you follow NCERT or Ganita Manjari, this resource supports quick revision and effective exam preparation.

Table of Contents

• Key Concepts on Predicting What Comes Next: Exploring Sequences and Progressions for Class 9
• Important Questions on Sequences and Progressions for Class 9

Key Concepts on Predicting What Comes Next: Exploring Sequences and Progressions for Class 9

Memorise these formulas of sequences and progressions, then practice applying them in different contexts.

Important Questions on Sequences and Progressions for Class 9

Q1: The 10th term of the AP 3, 7, 11, 15,… is:

Solution: Using aₙ = a + (n – 1)d, with a = 3, d = 4, n = 10:

a₁₀ = 3 + (10 – 1) × 4 = 3 + 36 = 39

Q2: The common ratio of the GP 4, 12, 36, 108,… is:

Solution: r = a₂/a₁ = 12/4 = 3

Q3: If the nth term of a sequence is given by aₙ = 3n + 2, what are the first three terms?

Solution: a₁ = 3(1) + 2 = 5;   a₂ = 3(2) + 2 = 8;   a₃ = 3(3) + 2 = 11

First three terms: 5, 8, 11 (which also forms an AP with d = 3).

Q4: Find the number of terms in the AP: 7, 13, 19, … , 205.

Solution: Given: a = 7, d = 13 – 7 = 6, last term l = 205

l = a + (n – 1)d

205 = 7 + (n – 1) × 6 

⇒ 198 = (n – 1) × 6 

⇒ n – 1 = 33 

⇒ n = 34

There are 34 terms in this AP.

Q5: Is 200 a term of the AP 3, 7, 11, 15, …? Give reasons.

Solution: Given: a = 3, d = 4.

Assume 200 = aₙ

aₙ = a + (n – 1)d

200 = 3 + (n – 1) × 4 

⇒ 197 = (n – 1) × 4 

⇒ n – 1 = 49.25

Since n is not a natural number, 200 is not a term of the given AP.

Q6: The 5th and 8th terms of a GP are 32 and 256 respectively. Find the common ratio. 

Solution: Given: a₅ = ar⁴ = 32 and a₈ = ar⁷ = 256

Divide: a₈/a₅ = r³ = 256/32 = 8

⇒ r = ∛8 = 2

Q7: Find the sum of the first 20 natural numbers.

Solution: Natural numbers 1, 2, 3, … , 20 form an AP with a = 1, d = 1, n = 20.

Sₙ = n/2 × (a + l) 

= 20/2 × (1 + 20) 

= 10 × 21 = 210

Q8: Write the general term (nth term) of the sequence: 2, 6, 12, 20, 30, …

Solution: Given sequence is 2, 6, 12, 20, 30, …

Differences: 4, 6, 8, 10… (differences of differences = 2, constant). This is not an AP or GP. Look for a pattern:

2 = 1 × 2, 6 = 2 × 3, 12 = 3 × 4, 20 = 4 × 5, 30 = 5 × 6

The nth term is aₙ = n(n + 1). Check: a₁ = 1(2) = 2, a₅ = 5(6) = 30

Q9: A ball is dropped from a height of 80 m. Each time it hits the ground, it bounces back to 3/4 of the height from which it fell. Find the total distance travelled by the ball before it comes to rest.

Solution: Down: Ball falls 80 m for the first time.

Up+Down:

Then bounces up 60 m (= 80 × 3/4), falls 60 m, bounces up 45 m, falls 45 m… This is a GP with a = 60, r = 3/4 for the up-down pairs.

GP Sum:

Sum of all up-down distances = 2 × 60/(1 – 3/4) = 2 × 60/(1/4) = 2 × 240 = 480 m

Total: 80 (first fall) + 480 = 560 m

Q10: An AP consists of 37 terms. The sum of the three middle terms is 225 and the sum of the last three terms is 429. Find the AP.

Solution: number of terms = 37 terms 

⇒ middle term is a₁₉. 

Three middle terms: a₁₈, a₁₉, a₂₀. 

Their sum = 3a₁₉ = 225 

⇒ a₁₉ = 75.

Last three terms are a₃₅, a₃₆, a₃₇. 

Sum = 3a₃₆ = 429 

⇒ a₃₆ = 143.

Find d: a₃₆ – a₁₉ = 17d 

⇒ 143 – 75 = 17d 

⇒ 68 = 17d 

⇒ d = 4.

Find a:

a₁₉ = a + 18d ⇒ 75 = a + 72 ⇒ a = 3.

AP: 3, 7, 11, 15, … with first term 3 and common difference 4.

Q11: If the pth, qth, and rth terms of an AP are a, b, and c respectively, show that a(q – r) + b(r – p) + c(p – q) = 0.

Solution: Let:First term = A, common difference = D. 

Then a = A + (p – 1)D, b = A + (q – 1)D, and c = A + (r – 1)D.

Substitute:

a(q–r) + b(r–p) + c(p–q) = A[(q–r) + (r–p) + (p–q)] + D[(p–1)(q–r) + (q–1)(r–p) + (r–1)(p–q)]

First bracket = 0. Second bracket = (pq – pr – q + r) + (qr – pq – r + p) + (rp – qr – p + q) = 0.

Therefore the entire expression = 0. Proved.