Chapter 8: Predicting What Comes Next: Exploring Sequences and ProgressionsPage 179–180All 6 Questions on This Page

Class 9 Maths Chapter 8 Exercise Set 8.1 Solutions

Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 8 Exercise Set 8.1. Covers evaluating first terms of sequences from nth term formulas, determining specific terms, checking membership in a sequence, and working with recursive sequence definitions.

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Key Concepts for Exercise Set 8.1

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Sequence and nth Term (tₙ)

A sequence is an ordered list of numbers following a definite rule. The nth term, denoted by tₙ, provides an algebraic formula for the term at position n.

tₙ = f(n), where n ∈ {1, 2, 3, …}

Finding Terms by Substituting n

To find the first k terms of a sequence when tₙ is given, substitute integer values n = 1, 2, 3, … into the general nth term formula.

First term = t₁, Second term = t₂, Third term = t₃, …

Checking Sequence Membership

To determine whether a number is a term of a sequence, set tₙ equal to that number and solve for n. If n is a positive integer (1, 2, 3, …), the number belongs to the sequence.

Finding the Position n of a Given Term

When a value is known to belong to the sequence, solving the linear equation tₙ = value gives its exact position n.

5n − 3 = 607 ⇒ 5n = 610 ⇒ n = 122 (122nd term)

Recursive Rules for Sequences

A recursive formula defines each term using the term(s) immediately before it, starting from specified initial terms (base cases).

t₁ = a, tₙ₊₁ = tₙ + d, for n ≥ 1

Arithmetic Progression as a Recursive Sequence

When each term is obtained by adding a constant difference d to the previous term (tₙ₊₁ = tₙ + d), the sequence forms an Arithmetic Progression with common difference d.

t₁ = −5, d = 3 ⇒ tₙ = a + (n − 1)d

Sequences Involving Multiple Preceding Terms

Some recursive sequences depend on more than one preceding term (such as Tribonacci or Fibonacci-style sequences), requiring multiple initial values to get started.

Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃, for n ≥ 4

Question 1

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Find the first five terms of the sequence in which the nth term is given by:

(i)tₙ = 3n − 4
(ii)tₙ = 2 − 5n
(iii)tₙ = n² − 2n + 3, for n ≥ 1

Solution

(i) tₙ = 3n − 4:

Substitute n = 1, 2, 3, 4, 5:

t₁ = 3(1) − 4 = −1

t₂ = 3(2) − 4 = 2

t₃ = 3(3) − 4 = 5

t₄ = 3(4) − 4 = 8

t₅ = 3(5) − 4 = 11

Therefore, the first five terms are:

−1, 2, 5, 8, 11

(ii) tₙ = 2 − 5n:

Substitute n = 1, 2, 3, 4, 5:

t₁ = 2 − 5(1) = 2 − 5 = −3

t₂ = 2 − 5(2) = 2 − 10 = −8

t₃ = 2 − 5(3) = 2 − 15 = −13

t₄ = 2 − 5(4) = 2 − 20 = −18

t₅ = 2 − 5(5) = 2 − 25 = −23

Therefore, the first five terms are:

−3, −8, −13, −18, −23

(iii) tₙ = n² − 2n + 3:

Substitute n = 1, 2, 3, 4, 5:

t₁ = 1² − 2(1) + 3 = 2

t₂ = 2² − 2(2) + 3 = 3

t₃ = 3² − 2(3) + 3 = 6

t₄ = 4² − 2(4) + 3 = 11

t₅ = 5² − 2(5) + 3 = 18

Therefore, the first five terms are:

2, 3, 6, 11, 18

Answer:(i) −1, 2, 5, 8, 11 | (ii) −3, −8, −13, −18, −23 | (iii) 2, 3, 6, 11, 18

Question 2

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Find the 10th and 15th terms of the sequence: tₙ = 5n − 3, for n ≥ 1.

Solution

Given:

tₙ = 5n − 3

For the 10th term:

t₁₀ = 5(10) − 3

= 50 − 3

= 47

For the 15th term:

t₁₅ = 5(15) − 3

= 75 − 3

= 72

Therefore:

t₁₀ = 47

t₁₅ = 72

Answer:t₁₀ = 47 and t₁₅ = 72

Question 3

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Determine whether 97 and 172 are terms of the sequence: tₙ = 5n − 3, for n ≥ 1.

Solution

For 97:

5n − 3 = 97

5n = 100

n = 20

Since n = 20 is a positive integer, 97 is a term of the sequence.

Therefore:

97 = t₂₀

For 172:

5n − 3 = 172

5n = 175

n = 35

Since n = 35 is a positive integer, 172 is also a term of the sequence.

Therefore:

172 = t₃₅

Hence:

97 = t₂₀

172 = t₃₅

Answer:Both are terms: 97 is the 20th term (t₂₀ = 97), and 172 is the 35th term (t₃₅ = 172).

Question 4

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Which term of the sequence: tₙ = 5n − 3, for n ≥ 1 is 607?

Solution

We need:

tₙ = 607

So,

5n − 3 = 607

5n = 610

n = 122

Therefore:

607 is the 122nd term.

Answer:607 is the 122nd term (t₁₂₂ = 607).

Question 5

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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

Given:

t₁ = −5

Each term is obtained by adding 3 to the previous term:

t₂ = t₁ + 3 = −5 + 3 = −2

t₃ = t₂ + 3 = −2 + 3 = 1

t₄ = t₃ + 3 = 1 + 3 = 4

t₅ = t₄ + 3 = 4 + 3 = 7

Therefore, the first five terms are:

−5, −2, 1, 4, 7

To check whether 52 is a term, observe that the sequence is an AP with:

a = −5

d = 3

Using:

tₙ = a + (n − 1)d

For 52 to be a term:

−5 + (n − 1)3 = 52

3(n − 1) = 57

n − 1 = 19

n = 20

Since n = 20 is a positive integer, 52 is a term.

Therefore:

52 = t₂₀

So, 52 is the 20th term.

Answer:First five terms are −5, −2, 1, 4, 7; Yes, 52 is the 20th term (t₂₀ = 52).

Question 6

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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

Given:

T₁ = 1

T₂ = 2

T₃ = 4

Using the recursive rule:

T₄ = T₃ + T₂ + T₁ = 4 + 2 + 1 = 7

T₅ = T₄ + T₃ + T₂ = 7 + 4 + 2 = 13

T₆ = T₅ + T₄ + T₃ = 13 + 7 + 4 = 24

T₇ = T₆ + T₅ + T₄ = 24 + 13 + 7 = 44

T₈ = T₇ + T₆ + T₅ = 44 + 24 + 13 = 81

Therefore:

T₄ = 7

T₅ = 13

T₆ = 24

T₇ = 44

T₈ = 81

Answer:T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81

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💡Key Concepts for Exercise Set 8.1

Sequence and nth Term (tₙ)

Finding Terms by Substituting n

Checking Sequence Membership

Finding the Position n of a Given Term

Recursive Rules for Sequences

Arithmetic Progression as a Recursive Sequence

Sequences Involving Multiple Preceding Terms

Key Concepts for Exercise Set 8.1