Chapter 8 — Sequences and Series

Concept Insight:

A sequence defined by a formula \(a_n = f(n)\) generates terms by substituting natural numbers. Here, \(a_n = n(n+2)\) is a quadratic sequence.

Expanding: \(a_n = n^2 + 2n\), which indicates:

• Growth is non-linear
• Differences between terms increase steadily
• Second difference is constant → confirms quadratic nature

Solution Roadmap:

• Step 1: Substitute \(n = 1,2,3,4,5\)
• Step 2: Evaluate expression carefully
• Step 3: Observe pattern (growth trend)

Solution

Substitute values of \(n\) into \(a_n = n(n+2)\):

$$ \begin{aligned} a_1 &= 1(3) = 3 \\ a_2 &= 2(4) = 8 \\ a_3 &= 3(5) = 15 \\ a_4 &= 4(6) = 24 \\ a_5 &= 5(7) = 35 \end{aligned} $$

First five terms: \(3, 8, 15, 24, 35\)

Notice how the increase between terms grows: \(+5, +7, +9, +11\). This confirms a quadratic pattern.

Exam Relevance:

• CBSE Boards: Direct substitution-based questions are common (1–2 marks)
• JEE/NEET: Helps identify sequence type quickly
• Foundation Concept: Used later in AP, GP, and Series expansion problems
• Speed Skill: Reduces calculation time in MCQs

Concept Insight:

The sequence \(a_n = \dfrac{n}{n+1}\) is a rational sequence.

Rewrite: \[ a_n = 1 - \dfrac{1}{n+1} \]

• Each term is slightly less than 1
• The difference from 1 decreases as \(n\) increases
• This is an example of a sequence that converges to 1

Solution Roadmap:

• Step 1: Substitute \(n = 1,2,3,4,5\)
• Step 2: Simplify each fraction carefully
• Step 3: Observe how values approach 1

Solution

Substitute values of \(n\) into \(a_n = \dfrac{n}{n+1}\):

$$ \begin{aligned} a_1 &= \dfrac{1}{2} \\ a_2 &= \dfrac{2}{3} \\ a_3 &= \dfrac{3}{4} \\ a_4 &= \dfrac{4}{5} \\ a_5 &= \dfrac{5}{6} \end{aligned} $$

First five terms: \[ \dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, \dfrac{4}{5}, \dfrac{5}{6} \]

The graph shows values increasing and getting closer to a fixed level (horizontal line). This indicates the sequence is approaching 1 but never reaching it.

Exam Relevance:

• CBSE Boards: Direct substitution + simplification (easy marks)
• JEE/NEET: Frequently used in limit-based questions
• Concept Builder: Introduces idea of convergence
• Advanced Use: Helps in understanding series like harmonic and telescoping series

Concept Insight:

The sequence \(a_n = 2^n\) is a geometric sequence where each term is obtained by multiplying the previous term by a constant ratio.

• First term: \(a_1 = 2\)
• Common ratio: \(r = 2\)
• Type: Exponential Growth

General form of GP: \(a_n = a \cdot r^{n-1}\) Here it matches with \(a = 2\), \(r = 2\)

Solution Roadmap:

• Step 1: Substitute \(n = 1,2,3,4,5\)
• Step 2: Evaluate powers of 2
• Step 3: Observe doubling pattern

Solution

Substitute values of \(n\) into \(a_n = 2^n\):

$$ \begin{aligned} a_1 &= 2^1 = 2 \\ a_2 &= 2^2 = 4 \\ a_3 &= 2^3 = 8 \\ a_4 &= 2^4 = 16 \\ a_5 &= 2^5 = 32 \end{aligned} $$

First five terms: \[ 2, 4, 8, 16, 32 \]

The spacing between terms increases rapidly, showing exponential growth. Each step multiplies the previous term by 2.

Exam Relevance:

• CBSE Boards: Direct GP identification + term writing
• JEE/NEET: Core base for exponential and GP sum problems
• High Weightage Concept: Used in growth/decay models
• Speed Trick: Recognize powers instantly (2,4,8,16…)

Concept Insight:

The sequence \(a_n = \dfrac{2n-3}{6}\) is a linear function of \(n\), hence it represents an Arithmetic Progression (AP).

Rewrite: \[ a_n = \frac{2n}{6} - \frac{3}{6} = \frac{n}{3} - \frac{1}{2} \]

• First term: \(a_1 = -\dfrac{1}{6}\)
• Common difference: \(d = \dfrac{1}{3}\)
• Linear growth → constant increase

Solution Roadmap:

• Step 1: Substitute \(n = 1,2,3,4,5\)
• Step 2: Simplify fractions carefully
• Step 3: Verify constant difference

Solution

Substitute values of \(n\) into \(a_n = \dfrac{2n-3}{6}\):

$$ \begin{aligned} a_1 &= \dfrac{-1}{6} \\ a_2 &= \dfrac{1}{6} \\ a_3 &= \dfrac{3}{6} = \dfrac{1}{2} \\ a_4 &= \dfrac{5}{6} \\ a_5 &= \dfrac{7}{6} \end{aligned} $$

First five terms: \[ -\dfrac{1}{6},\ \dfrac{1}{6},\ \dfrac{1}{2},\ \dfrac{5}{6},\ \dfrac{7}{6} \]

The equal spacing between points confirms a constant difference, which is the defining property of an arithmetic sequence.

Exam Relevance:

• CBSE Boards: Identifying AP from formula is frequently asked
• JEE/NEET: Used in nth term and sum of AP problems
• Concept Skill: Converting algebraic form → AP form is key
• Trap Alert: Students often miss rewriting into standard AP form

Concept Insight:

The sequence \(a_n = (-1)^{n-1} \cdot 5^{n+1}\) combines two important behaviors:

• Alternating Sign: \( (-1)^{n-1} \) makes terms positive for odd \(n\), negative for even \(n\)
• Exponential Growth: \(5^{n+1}\) increases rapidly

This is an alternating geometric-type sequence where magnitude grows while sign flips.

Solution Roadmap:

• Step 1: Determine sign using \( (-1)^{n-1} \)
• Step 2: Compute \(5^{n+1}\)
• Step 3: Multiply sign with magnitude

Solution

Substitute values of \(n\):

$$ \begin{aligned} a_1 &= (+1)\cdot 5^2 = 25 \\ a_2 &= (-1)\cdot 5^3 = -125 \\ a_3 &= (+1)\cdot 5^4 = 625 \\ a_4 &= (-1)\cdot 5^5 = -3125 \\ a_5 &= (+1)\cdot 5^6 = 15625 \end{aligned} $$

First five terms: \[ 25,\ -125,\ 625,\ -3125,\ 15625 \]

The graph shows a zig-zag pattern: values alternate above and below the axis, while magnitude increases rapidly due to exponential growth.

Exam Relevance:

• CBSE Boards: Testing substitution + sign handling
• JEE/NEET: Frequently appears in alternating series and convergence questions
• Concept Skill: Separating sign pattern and magnitude is critical
• Trap Alert: Students often ignore \( (-1)^{n-1} \) and lose marks

Concept Insight:

The sequence \(a_n = n\dfrac{n^2+5}{4}\) is a cubic sequence.

Rewrite: \[ a_n = \frac{n^3 + 5n}{4} \]

• Highest power is \(n^3\) → cubic growth
• Growth accelerates faster than quadratic or linear sequences
• Values can be fractions or integers depending on divisibility

Solution Roadmap:

• Step 1: Substitute \(n = 1,2,3,4,5\)
• Step 2: Compute \(n^2 + 5\)
• Step 3: Multiply by \(n\) and divide by 4
• Step 4: Simplify fractions

Solution

Substitute values of \(n\):

$$ \begin{aligned} a_1 &= \dfrac{6}{4} = \dfrac{3}{2} \\ a_2 &= \dfrac{18}{4} = \dfrac{9}{2} \\ a_3 &= \dfrac{42}{4} = \dfrac{21}{2} \\ a_4 &= \dfrac{84}{4} = 21 \ a_5 &= \dfrac{150}{4} = \dfrac{75}{2} \end{aligned} $$

First five terms: \[ \dfrac{3}{2},\ \dfrac{9}{2},\ \dfrac{21}{2},\ 21,\ \dfrac{75}{2} \]

The curve rises increasingly steeply, indicating cubic growth. Differences between terms increase rapidly.

Exam Relevance:

• CBSE Boards: Direct substitution + simplification
• JEE/NEET: Helps identify polynomial degree from sequence
• Concept Skill: Recognizing cubic vs quadratic patterns
• Trap Alert: Students often mis-handle fraction simplification

Concept Insight:

The expression \(a_n = 4n - 3\) represents an Arithmetic Progression (AP).

• First term: \(a_1 = 4(1) - 3 = 1\)
• Common difference: \(d = 4\)
• General form: linear in \(n\) → constant increase

Instead of listing all terms up to 17 or 24, we directly use the formula to compute required terms.

Solution Roadmap:

• Step 1: Identify formula \(a_n = 4n - 3\)
• Step 2: Substitute \(n = 17\) and \(n = 24\)
• Step 3: Simplify arithmetic carefully

Solution

$$ \begin{aligned} a_{17} &= 4 \times 17 - 3 \\&= 68 - 3 \\&= 65 \\\\ a_{24} &= 4 \times 24 - 3 \\&= 96 - 3 \\&= 93 \end{aligned} $$

Final Answer: \[ a_{17} = 65,\quad a_{24} = 93 \]

The straight-line pattern confirms a constant difference of 4, which defines an arithmetic progression.

Exam Relevance:

• CBSE Boards: Direct substitution (very common 1–2 mark question)
• JEE/NEET: Faster than listing terms → saves time
• Concept Skill: Recognize AP instantly from linear expression
• Trap Alert: Students sometimes miscalculate large values (like 24)

Concept Insight:

The sequence \(a_n = \dfrac{n^2}{2^n}\) combines:

• Polynomial Growth: \(n^2\)
• Exponential Growth: \(2^n\)

Since exponential growth dominates polynomial growth, the sequence decreases and approaches 0 as \(n\) increases.

Solution Roadmap:

• Step 1: Substitute \(n = 7\)
• Step 2: Compute numerator \(7^2\)
• Step 3: Compute denominator \(2^7\)
• Step 4: Simplify fraction

Solution

$$ \begin{aligned} a_7 &= \dfrac{7^2}{2^7} \\ &= \dfrac{49}{128} \end{aligned} $$

Final Answer: \[ a_7 = \dfrac{49}{128} \]

The graph shows values decreasing as \(n\) increases, illustrating how exponential growth in the denominator dominates.

Exam Relevance:

• CBSE Boards: Direct substitution + simplification
• JEE/NEET: Core concept in limits (\(\lim n^2/2^n = 0\))
• Concept Skill: Comparing polynomial vs exponential growth
• Trap Alert: Students sometimes compute \(2^7\) incorrectly

Concept Insight:

The sequence \(a_n = (-1)^{n-1} n^3\) combines:

• Alternating Sign: \( (-1)^{n-1} \)
• Cubic Growth: \(n^3\)

Rule:

• Odd \(n\) → positive term
• Even \(n\) → negative term

Solution Roadmap:

• Step 1: Identify parity of \(n\)
• Step 2: Determine sign using \( (-1)^{n-1} \)
• Step 3: Compute \(n^3\)
• Step 4: Multiply sign and value

Solution

$$ \begin{aligned} a_9 &= (-1)^{9-1} \cdot 9^3 \\ &= (-1)^8 \cdot 729 \\ &= 1 \cdot 729 \\ &= 729 \end{aligned} $$

Final Answer: \[ a_9 = 729 \]

The graph shows alternating signs (zig-zag) while magnitudes increase rapidly due to cubic growth.

Exam Relevance:

• CBSE Boards: Tests substitution + sign logic
• JEE/NEET: Important in alternating series and sequence behavior
• Concept Skill: Separating sign pattern from magnitude
• Trap Alert: Mistakes in evaluating \( (-1)^{n-1} \) are very common

Concept Insight:

The sequence \(a_n = \dfrac{n(n-2)}{n+3}\) is a rational function.

Simplify using division: \[ a_n = n - 5 + \frac{15}{n+3} \]

• Behaves like a linear expression \(n - 5\)
• The extra term \(\frac{15}{n+3}\) becomes very small for large \(n\)
• Hence the sequence shows near-linear growth

Solution Roadmap:

• Step 1: Substitute \(n = 20\)
• Step 2: Compute numerator and denominator
• Step 3: Simplify fraction

Solution

$$ \begin{aligned} a_{20} &= \dfrac{20(20-2)}{20+3} \\ &= \dfrac{20 \times 18}{23} \\ &= \dfrac{360}{23} \end{aligned} $$

Final Answer: \[ a_{20} = \dfrac{360}{23} \]

The graph appears almost linear, confirming that the sequence behaves like \(n - 5\) for large \(n\), with a small correction term.

Exam Relevance:

• CBSE Boards: Direct substitution + simplification
• JEE/NEET: Important for limit and asymptotic behavior
• Concept Skill: Simplifying rational expressions
• Trap Alert: Mistakes in multiplication or denominator handling

Concept Insight:

This is a recursive sequence, where each term depends on the previous one.

• Start value: \(a_1 = 3\)
• Rule: multiply previous term by 3, then add 2

Such sequences often show exponential-type growth due to repeated multiplication.

Solution Roadmap:

• Step 1: Start from \(a_1\)
• Step 2: Apply recurrence repeatedly
• Step 3: Compute up to \(a_5\)
• Step 4: Write the series (sum form)

Solution

$$ \begin{aligned} a_1 &= 3 \\ a_2 &= 3(3) + 2 = 11 \\ a_3 &= 3(11) + 2 = 35 \\ a_4 &= 3(35) + 2 = 107 \\ a_5 &= 3(107) + 2 = 323 \end{aligned} $$

First five terms: \[ 3,\ 11,\ 35,\ 107,\ 323 \]

Corresponding series: \[ 3 + 11 + 35 + 107 + 323 \]

The rapid rise shows recursive amplification — each step multiplies growth.

Exam Relevance:

• CBSE Boards: Writing terms from recurrence is common
• JEE/NEET: Basis for recurrence relations and series
• Concept Skill: Iterative thinking (step-by-step dependency)
• Trap Alert: Students often skip steps and make calculation errors

Concept Insight:

This is a recursive sequence where each term is obtained by dividing the previous term by \(n\).

• Starting value: \(a_1 = -1\)
• Each step reduces magnitude (division by increasing \(n\))
• All terms remain negative

Pattern recognition: \[ a_n = -\frac{1}{n!} \] This connects the sequence to factorials.

Solution Roadmap:

• Step 1: Start from \(a_1\)
• Step 2: Divide successively by \(2,3,4,5\)
• Step 3: Observe factorial pattern
• Step 4: Write sequence and corresponding series

Solution

$$ \begin{aligned} a_1 &= -1 \\ a_2 &= -\dfrac{1}{2} \\ a_3 &= -\dfrac{1}{6} \\ a_4 &= -\dfrac{1}{24} \\ a_5 &= -\dfrac{1}{120} \end{aligned} $$

First five terms: \[ -1,\ -\frac{1}{2},\ -\frac{1}{6},\ -\frac{1}{24},\ -\frac{1}{120} \]

Corresponding series: \[ -1 - \frac{1}{2} - \frac{1}{6} - \frac{1}{24} - \frac{1}{120} \]

The terms move closer to zero, showing factorial decay. Magnitude decreases rapidly as \(n!\) grows.

Exam Relevance:

• CBSE Boards: Recursive computation + pattern recognition
• JEE/NEET: Strong connection to series expansions (e.g., \(e^x\))
• Concept Skill: Identifying factorial structure
• Trap Alert: Students often miss factorial pattern

Concept Insight:

This is a recursive sequence with two initial terms.

• First two terms are equal: \(a_1 = a_2 = 2\)
• After that, each term decreases by 1

From \(n \geq 3\), the sequence behaves like an Arithmetic Progression (AP) with common difference \(d = -1\).

Solution Roadmap:

• Step 1: Start from given initial terms
• Step 2: Apply recurrence \(a_n = a_{n-1} - 1\)
• Step 3: Generate terms up to \(a_5\)
• Step 4: Write corresponding series

Solution

$$ \begin{aligned} a_1 &= 2 \\ a_2 &= 2 \\ a_3 &= 1 \\ a_4 &= 0 \\ a_5 &= -1 \end{aligned} $$

First five terms: \[ 2,\ 2,\ 1,\ 0,\ -1 \]

Corresponding series: \[ 2 + 2 + 1 + 0 - 1 \]

After the first two equal terms, the sequence decreases uniformly, showing linear decay with common difference \(-1\).

Exam Relevance:

• CBSE Boards: Recursive sequence with multiple initial conditions
• JEE/NEET: Important for identifying hidden AP behavior
• Concept Skill: Transition from recursion → AP recognition
• Trap Alert: Students often assume AP from start (ignore first two terms)

Concept Insight:

The Fibonacci sequence is a recursive sequence where each term is the sum of the previous two.

• Start: \(1, 1\)
• Growth: additive (not multiplicative)
• Key property: ratios of consecutive terms approach a constant

This constant is the Golden Ratio: \[ \phi \approx 1.618 \]

Solution Roadmap:

• Step 1: Generate terms up to \(a_6\)
• Step 2: Compute ratios \(\frac{a_{n+1}}{a_n}\)
• Step 3: Observe convergence pattern

Solution

$$ \begin{aligned} a_1 &= 1,\quad a_2 = 1 \\ a_3 &= 2 \\ a_4 &= 3 \\ a_5 &= 5 \\ a_6 &= 8 \end{aligned} $$

$$ \begin{aligned} \frac{a_2}{a_1} &= 1 \\ \frac{a_3}{a_2} &= 2 \\ \frac{a_4}{a_3} &= \frac{3}{2} \\ \frac{a_5}{a_4} &= \frac{5}{3} \\ \frac{a_6}{a_5} &= \frac{8}{5} \end{aligned} $$

Ratios: \[ 1,\ 2,\ \frac{3}{2},\ \frac{5}{3},\ \frac{8}{5} \]

The ratios oscillate and settle toward a fixed value — the Golden Ratio.

Exam Relevance:

• CBSE Boards: Recurrence + ratio evaluation
• JEE/NEET: Golden ratio and limits are frequently tested
• Concept Skill: Understanding convergence behavior
• Advanced Link: Appears in algebra, number theory, and nature patterns