Class 9 - Ganita Manjari Chapter 8 - Part 24
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Class 9 Ganita Manjari Chapter 8: Sequences and Progressions
Part 24
CHAIN REACTIONS End of Chapter - Recursive Chains \(t_n = 2t_{n-1} + 3\) ×2 + 3 ×2 + 3 1 5 13 The Past Dictates the Future
As we approach the finale of Chapter 8, we step beyond standard APs and GPs to tackle Advanced Recursive Chains.

In these sequences, the formula does not just add a difference or multiply by a ratio—it does both, or it relies on multiple past terms simultaneously! These chains force you to build the sequence methodically, step-by-step, perfectly executing the mathematical operations dictated by the recursion.
1 Multi-Operation Recursion
Ganita Manjari - Pg 195
End of Chapter Exercise: A sequence is defined recursively by \(t_1 = 3\) and \(t_n = 2t_{n-1} + 1\) for \(n > 1\). Find the first four terms of this sequence.
Concept:
1
A recursive rule like \(t_n = 2t_{n-1} + 1\) means "to find the next term, multiply the previous term by 2, then add 1".
2
You must calculate these strictly in order. You cannot find \(t_4\) without first knowing \(t_3\).
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Hint:
1
\(t_1 = 3\).
2
\(t_2 = 2(t_1) + 1 = 2(3) + 1 = 7\).
3
Now use 7 to find \(t_3\)!
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Mistakes to Avoid:
!
Order of Operations: Always multiply before you add. Do not evaluate \(2(3) + 1\) as \(2 \times 4 = 8\). It must be \(6 + 1 = 7\).
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Solution:
Given: \(t_1 = 3\) and \(t_n = 2t_{n-1} + 1\)
\(t_1 = 3\)
\(t_2 = 2(3) + 1 = 6 + 1 = 7\)
\(t_3 = 2(7) + 1 = 14 + 1 = 15\)
\(t_4 = 2(15) + 1 = 30 + 1 = 31\)

Ans: 3, 7, 15, 31
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2 Multi-Term Dependence
Ganita Manjari - Pg 195
End of Chapter Exercise: A sequence is defined by \(t_1 = 1\), \(t_2 = 2\), and the recursive relation \(t_n = t_{n-1} \times t_{n-2}\) for \(n \ge 3\). Find the 5th term of the sequence.
Concept:
1
This is a product-based sequence similar to Fibonacci. To find the next term, multiply the two immediately preceding terms together.
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Hint:
1
\(t_3 = t_2 \times t_1 = 2 \times 1 = 2\).
2
\(t_4 = t_3 \times t_2\). Then find \(t_5\).
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Mistakes to Avoid:
!
Adding instead of multiplying: Do not compute \(t_3 = 1 + 2 = 3\). The formula explicitly dictates multiplication (\(\times\)).
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Solution:
Given: \(t_1 = 1\), \(t_2 = 2\)
\(t_3 = t_2 \times t_1 = 2 \times 1 = 2\)
\(t_4 = t_3 \times t_2 = 2 \times 2 = 4\)
\(t_5 = t_4 \times t_3 = 4 \times 2 = 8\)

Ans: The 5th term is 8.
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★ Algebraic Recursive Chains
Ganita Manjari - Pg 195
End of Chapter Exercise (*): Let a sequence be defined by \(t_n = t_{n-1} + t_{n-2}\). If \(t_1 = a\) and \(t_2 = b\), express the 5th term (\(t_5\)) entirely in terms of the variables \(a\) and \(b\).
Concept:
1
This is the classic Fibonacci recursive structure, but applied algebraically.
2
You must algebraically add the previous two expressions to generate the next one, combining like terms (\(a\)'s and \(b\)'s) at each step.
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Hint:
1
\(t_3 = t_2 + t_1 = b + a = a + b\).
2
\(t_4 = t_3 + t_2 = (a+b) + b = a + 2b\).
3
Now add \(t_4\) and \(t_3\) to find \(t_5\).
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Mistakes to Avoid:
!
Treating variables as constants: Do not attempt to assign numbers to \(a\) and \(b\). The answer must remain as an algebraic expression.
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Solution:
1. Define base terms:
\(t_1 = a\)
\(t_2 = b\)

2. Apply recursive rule \(t_n = t_{n-1} + t_{n-2}\):
\(t_3 = b + a = a + b\)
\(t_4 = (a+b) + b = a + 2b\)
\(t_5 = (a+2b) + (a+b) = 2a + 3b\)

Ans: \(2a + 3b\)
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Select the correct option:
If a sequence is defined recursively as \(t_n = t_{n-1} + 5\), what mathematical class does this sequence belong to?
Concept:
1
A recursive rule that simply adds a constant \(d\) is the fundamental definition of an AP: \(t_n = t_{n-1} + d\).
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Hint:
1
If a pattern goes \(+5, +5, +5\), what kind of progression is that?
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Mistakes to Avoid:
!
Assuming all recursive rules are complex: While recursion sounds advanced, simple recursive addition is just an AP in disguise.
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Solution:
The rule \(t_n = t_{n-1} + 5\) means every step grows by a common difference of 5. This forms an Arithmetic Progression.
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Select the correct option:
If a sequence is defined recursively as \(t_n = 3 \cdot t_{n-1}\), what mathematical class does this sequence belong to?
Concept:
1
A recursive rule that multiplies the previous term by a constant ratio \(r\) is the fundamental definition of a GP: \(t_n = r \cdot t_{n-1}\).
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Hint:
1
If a pattern is formed by multiplying by 3 repeatedly, what kind of progression is it?
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Mistakes to Avoid:
!
Assuming all GPs use powers: While the *explicit* formula uses exponents (\(r^n\)), the *recursive* formula uses simple multiplication (\(r \cdot t_{n-1}\)).
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Solution:
The rule \(t_n = 3 \cdot t_{n-1}\) establishes a common ratio of 3 between consecutive terms. This is a Geometric Progression.
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Select the correct option:
If a sequence is defined by \(t_1 = 5\) and \(t_n = -t_{n-1}\) for \(n > 1\), what is the value of the 4th term (\(t_4\))?
Concept:
1
This is an alternating sequence (a GP with \(r=-1\)). It simply flips the sign of the previous number.
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Hint:
1
\(t_1 = 5\)
2
\(t_2 = -(5) = -5\)
3
\(t_3 = -(-5) = 5\). Keep going.
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Mistakes to Avoid:
!
Losing track of the parity: Even terms (2, 4, 6) will be negative. Odd terms (1, 3, 5) will be positive.
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Solution:
\(t_1 = 5\)
\(t_2 = -5\)
\(t_3 = 5\)
\(t_4 = -5\)
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Select the correct option:
If you are given the recursive formula \(t_n = t_{n-1} + t_{n-2}\), why is it impossible to calculate \(t_{100}\) if you are only provided with the value of \(t_1 = 5\)?
Concept:
1
A recursive sequence must define enough base cases to satisfy the formula. If a formula looks back \(k\) steps, it must provide \(k\) starting terms.
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Hint:
1
Try to find \(t_3\). It equals \(t_2 + t_1\). You know \(t_1 = 5\), but what is \(t_2\)? You cannot proceed!
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Mistakes to Avoid:
!
Assuming \(t_2\) is 0 or 1: You cannot assume base cases in mathematics. If they are missing, the sequence is undefined.
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Solution:
Because the formula \(t_n = t_{n-1} + t_{n-2}\) requires the two immediately preceding terms, it is impossible to calculate \(t_3\) without knowing both \(t_1\) and \(t_2\). The chain is broken before it even starts.
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Select the correct option:
In a purely recursive sequence (like \(t_n = 2t_{n-1} + 3\)), what is the primary disadvantage compared to an explicit sequence formula (like \(t_n = 3 \cdot 2^n\))?
Concept:
1
Explicit formulas act like a "time machine", allowing you to jump to any position \(n\) instantly.
2
Recursive formulas act like "stepping stones", forcing you to cross every single stone to reach your destination.
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Hint:
1
If I want \(t_{1000}\) using \(t_n = 2t_{n-1}\), what do I need? I need \(t_{999}\). How do I get that? I need \(t_{998}\dots\)
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Mistakes to Avoid:
!
Assuming recursion is useless: While computationally heavy for large numbers, recursion is the foundation of computer science, programming loops, and natural fractals. It is essential, even if it lacks the jumping power of explicit algebra!
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Solution:
The major disadvantage of pure recursion is its dependency chain. You cannot evaluate a term without evaluating its prerequisite history.
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Chapter 8: Sequences & Progressions

Part 1: Intro to Sequences & Patterns Ganita Manjari Pgs 174-175 Part 2: Explicit Rules & Predicting Terms Ganita Manjari Pgs 176-177 Part 3: Recursive Rules & Fibonacci Ganita Manjari Pgs 178-179 Part 4: Exercise 8.1 - Mastery (Q1-Q3) Ganita Manjari Pg 179 Part 5: Exercise 8.1 - Recursive (Q4-Q6) Ganita Manjari Pg 179-180 Part 6: Intro to AP & Common Difference Ganita Manjari Pgs 180-182 Part 7: AP Formulas & Taxi Fare Math Ganita Manjari Pgs 182-183 Part 8: Sum of First n Natural Numbers Ganita Manjari Pgs 183-185 Part 9: Exercise 8.2 - AP Basics (Q1-Q3) Ganita Manjari Pg 185 Part 10: Exercise 8.2 - Missing Terms (Q4-Q5) Ganita Manjari Pgs 185-186 Part 11: Exercise 8.2 - AP Word Problems Ganita Manjari Pg 186 Part 12: Intro to GP & Common Ratios Ganita Manjari Pgs 186-188 Part 13: Fun with Fractals: SierpiΕ„ski Ganita Manjari Pgs 188-190 Part 14: GP Graphs & Bouncing Balls Ganita Manjari Pgs 190-193 Part 15: Exercise 8.3 - GP Basics Ganita Manjari Pg 193 Part 16: Exercise 8.3 - Rebounds & Roots Ganita Manjari Pgs 193-194 Part 17: Exercise 8.3 - SierpiΕ„ski Carpet Ganita Manjari Pg 194 Part 18: End of Chapter - AP Logic Ganita Manjari Pgs 194-195 Part 19: End of Chapter - Divisibility Ganita Manjari Pg 195 Part 20: End of Chapter - Consecutive Sums Ganita Manjari Pg 195 Part 21: End of Chapter - Dual Equations Ganita Manjari Pg 195 Part 22: End of Chapter - GP Proofs Ganita Manjari Pg 195 Part 23: End of Chapter - Complex Logic Ganita Manjari Pg 195 Part 24: End of Chapter - Recursive Chains Ganita Manjari Pg 195 Part 25: Chapter Summary & Finale Ganita Manjari Pg 196
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