2.3 Sequence and series

Detailed Theory: Sequences and Series

1. Basic Concepts and Definitions

1.1 Sequence

A sequence is a function whose domain is the set of natural numbers (or a subset of it).

Notation: $\{a_n\}$ or $a_1, a_2, a_3, \ldots$

$a_n$ is called the $n$-th term or general term of the sequence.

Example:

The sequence of natural numbers: $1, 2, 3, 4, 5, \ldots$ with $a_n = n$

1.2 Series

A series is the sum of the terms of a sequence.

For the sequence $\{a_n\}$, the corresponding series is:

$S_n = a_1 + a_2 + a_3 + \cdots + a_n = \sum_{k=1}^{n} a_k$

$S_n$ is called the $n$-th partial sum of the series.

1.3 Finite and Infinite Sequences/Series

Finite Sequence: Has a limited number of terms.

Example: $2, 4, 6, 8, 10$ (5 terms)

Infinite Sequence: Has infinitely many terms.

Example: $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$ with $a_n = \frac{1}{n}$

Finite Series: Sum of finite sequence terms.

Infinite Series: Sum of infinite sequence terms.

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2. Types of Sequences

2.1 Arithmetic Sequence (Arithmetic Progression)

A sequence where the difference between consecutive terms is constant.

Definition: $a_{n+1} - a_n = d$ for all $n$, where $d$ is the common difference.

General Form: $a, a+d, a+2d, a+3d, \ldots$

$n$-th term: $a_n = a + (n-1)d$

Example: $3, 7, 11, 15, 19, \ldots$ where $a=3$, $d=4$

$a_5 = 3 + (5-1) \times 4 = 3 + 16 = 19$

2.2 Geometric Sequence (Geometric Progression)

A sequence where the ratio between consecutive terms is constant.

Definition: $\frac{a_{n+1}}{a_n} = r$ for all $n$, where $r$ is the common ratio.

General Form: $a, ar, ar^2, ar^3, \ldots$

$n$-th term: $a_n = ar^{n-1}$

Example: $2, 6, 18, 54, \ldots$ where $a=2$, $r=3$

$a_4 = 2 \times 3^{3} = 2 \times 27 = 54$

2.3 Harmonic Sequence (Harmonic Progression)

A sequence where the reciprocals of the terms form an arithmetic progression.

Definition: $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots$ is an AP.

Example: $\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \ldots$ is an HP because $2, 4, 6, 8, \ldots$ is an AP.

Note: There's no simple formula for the $n$-th term of HP. Convert to AP first.

2.4 Fibonacci Sequence

A sequence where each term is the sum of the two preceding terms.

Definition: $a_1 = 1$, $a_2 = 1$, $a_n = a_{n-1} + a_{n-2}$ for $n \geq 3$

Sequence: $1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots$

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3. Arithmetic Progression (AP) - Detailed Study

3.1 Basic Properties

For an AP with first term $a$ and common difference $d$:

1. $n$-th term: $a_n = a + (n-1)d$

2. Sum of first $n$ terms:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Alternative form: $S_n = \frac{n}{2}(a + a_n)$

3. Arithmetic Mean (AM): For three terms in AP: $a-d, a, a+d$

For $n$ terms in AP, the arithmetic mean is:

$\text{AM} = \frac{a_1 + a_2 + \cdots + a_n}{n} = \frac{S_n}{n}$

3.2 Important Results for AP

1. If $a, b, c$ are in AP, then:

$b = \frac{a+c}{2}$ ($b$ is the arithmetic mean of $a$ and $c$)

$2b = a + c$

2. $m$-th term from beginning: $a_m = a + (m-1)d$

3. $m$-th term from end: For an AP with $n$ terms, the $m$-th term from end = $(n-m+1)$-th term from beginning

4. Sum of terms equidistant from ends:

For an AP: $a_1 + a_n = a_2 + a_{n-1} = a_3 + a_{n-2} = \cdots$

5. Selection of terms in AP:

Three terms: $a-d, a, a+d$

Four terms: $a-3d, a-d, a+d, a+3d$

Five terms: $a-2d, a-d, a, a+d, a+2d$

3.3 Solved Examples on AP

Example 1: Find the 20th term of AP: $3, 7, 11, 15, \ldots$

Solution:

Here $a=3$, $d=7-3=4$

$a_{20} = a + 19d = 3 + 19 \times 4 = 3 + 76 = 79$

Example 2: Find the sum of first 30 terms of AP: $5, 9, 13, 17, \ldots$

Solution:

$a=5$, $d=4$, $n=30$

$S_{30} = \frac{30}{2}[2\times5 + (30-1)\times4] = 15[10 + 29\times4]$

$= 15[10 + 116] = 15 \times 126 = 1890$

Example 3: If the sum of first $n$ terms of an AP is $3n^2 + 5n$, find the 25th term.

Solution:

$S_n = 3n^2 + 5n$

We know: $a_n = S_n - S_{n-1}$

So $a_n = [3n^2 + 5n] - [3(n-1)^2 + 5(n-1)]$

$= [3n^2 + 5n] - [3(n^2 - 2n + 1) + 5n - 5]$

$= [3n^2 + 5n] - [3n^2 - 6n + 3 + 5n - 5]$

$= [3n^2 + 5n] - [3n^2 - n - 2]$

$= 3n^2 + 5n - 3n^2 + n + 2$

$= 6n + 2$

So $a_{25} = 6 \times 25 + 2 = 150 + 2 = 152$

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4. Geometric Progression (GP) - Detailed Study

4.1 Basic Properties

For a GP with first term $a$ and common ratio $r$:

1. $n$-th term: $a_n = ar^{n-1}$

2. Sum of first $n$ terms:

If $r \neq 1$: $S_n = \frac{a(1 - r^n)}{1 - r}$ or $S_n = \frac{a(r^n - 1)}{r - 1}$

If $r = 1$: $S_n = na$

3. Sum of infinite GP: For $|r| < 1$:

$S_{\infty} = \frac{a}{1 - r}$

4. Geometric Mean (GM): For three terms in GP: $\frac{a}{r}, a, ar$

For $n$ positive numbers, the geometric mean is:

$\text{GM} = \sqrt[n]{a_1 \cdot a_2 \cdots a_n}$

4.2 Important Results for GP

1. If $a, b, c$ are in GP, then:

$b^2 = ac$ ($b$ is the geometric mean of $a$ and $c$)

$b = \sqrt{ac}$ (for positive numbers)

2. Product of terms equidistant from ends:

For a GP: $a_1 \cdot a_n = a_2 \cdot a_{n-1} = a_3 \cdot a_{n-2} = \cdots$

3. Product of first $n$ terms of GP:

$P_n = a^n r^{\frac{n(n-1)}{2}}$

4. Selection of terms in GP:

Three terms: $\frac{a}{r}, a, ar$

Four terms: $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$

Five terms: $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$

4.3 Solved Examples on GP

Example 1: Find the 10th term of GP: $2, 6, 18, 54, \ldots$

Solution:

Here $a=2$, $r=\frac{6}{2}=3$

$a_{10} = ar^{9} = 2 \times 3^9 = 2 \times 19683 = 39366$

Example 2: Find the sum of first 8 terms of GP: $3, 6, 12, 24, \ldots$

Solution:

$a=3$, $r=2$, $n=8$

Since $r>1$, use: $S_n = \frac{a(r^n - 1)}{r - 1}$

$S_8 = \frac{3(2^8 - 1)}{2 - 1} = \frac{3(256 - 1)}{1} = 3 \times 255 = 765$

Example 3: Find the sum to infinity: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$

Solution:

$a=1$, $r=\frac{1/2}{1}=\frac{1}{2}$

Since $|r|<1$, infinite sum exists:

$S_{\infty} = \frac{a}{1-r} = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$

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5. Harmonic Progression (HP) - Detailed Study

5.1 Basic Properties

A sequence is a harmonic progression if the reciprocals of its terms form an arithmetic progression.

Definition: $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots$ is an AP.

Harmonic Mean (HM): For two numbers $a$ and $b$:

$\text{HM} = \frac{2ab}{a+b}$

For $n$ numbers $a_1, a_2, \ldots, a_n$:

$\text{HM} = \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}$

5.2 Relationship between AP, GP, HP

For two positive numbers $a$ and $b$:

Arithmetic Mean: $\text{AM} = \frac{a+b}{2}$

Geometric Mean: $\text{GM} = \sqrt{ab}$

Harmonic Mean: $\text{HM} = \frac{2ab}{a+b}$

Important Inequality: For positive numbers:

$\text{AM} \geq \text{GM} \geq \text{HM}$

Equality holds if and only if $a = b$.

Proof of AM ≥ GM:

Consider $(\sqrt{a} - \sqrt{b})^2 \geq 0$

$a + b - 2\sqrt{ab} \geq 0$

$a + b \geq 2\sqrt{ab}$

$\frac{a+b}{2} \geq \sqrt{ab}$

So $\text{AM} \geq \text{GM}$

5.3 Solved Examples on HP

Example 1: Find the 8th term of HP: $\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots$

Solution:

The corresponding AP is: $3, 5, 7, 9, \ldots$

For this AP: $a=3$, $d=2$

8th term of AP: $a_8 = 3 + 7 \times 2 = 3 + 14 = 17$

So 8th term of HP = $\frac{1}{17}$

Example 2: Insert 3 harmonic means between 2 and 8.

Solution:

We need to insert 3 numbers between 2 and 8 such that all 5 numbers are in HP.

Let the HP be: $2, a, b, c, 8$

The corresponding AP is: $\frac{1}{2}, \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{8}$

In this AP: first term = $\frac{1}{2}$, fifth term = $\frac{1}{8}$

For AP: $a_5 = a_1 + 4d$

$\frac{1}{8} = \frac{1}{2} + 4d$

$4d = \frac{1}{8} - \frac{1}{2} = \frac{1-4}{8} = -\frac{3}{8}$

$d = -\frac{3}{32}$

Now: $\frac{1}{a} = \frac{1}{2} + d = \frac{1}{2} - \frac{3}{32} = \frac{16-3}{32} = \frac{13}{32} \Rightarrow a = \frac{32}{13}$

$\frac{1}{b} = \frac{1}{2} + 2d = \frac{1}{2} - \frac{6}{32} = \frac{16-6}{32} = \frac{10}{32} = \frac{5}{16} \Rightarrow b = \frac{16}{5}$

$\frac{1}{c} = \frac{1}{2} + 3d = \frac{1}{2} - \frac{9}{32} = \frac{16-9}{32} = \frac{7}{32} \Rightarrow c = \frac{32}{7}$

So harmonic means are: $\frac{32}{13}, \frac{16}{5}, \frac{32}{7}$

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6. Special Series and Summation Formulas

6.1 Sum of First $n$ Natural Numbers

$\sum_{k=1}^{n} k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$

6.2 Sum of Squares of First $n$ Natural Numbers

$\sum_{k=1}^{n} k^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$

6.3 Sum of Cubes of First $n$ Natural Numbers

$\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + 3^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$

Note: $(1 + 2 + 3 + \cdots + n)^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3$

6.4 Sum of First $n$ Odd Natural Numbers

$\sum_{k=1}^{n} (2k-1) = 1 + 3 + 5 + \cdots + (2n-1) = n^2$

6.5 Sum of First $n$ Even Natural Numbers

$\sum_{k=1}^{n} 2k = 2 + 4 + 6 + \cdots + 2n = n(n+1)$

6.6 Arithmetic-Geometric Progression (AGP)

A sequence where each term is the product of corresponding terms of an AP and a GP.

General Form: $a, (a+d)r, (a+2d)r^2, \ldots, [a+(n-1)d]r^{n-1}$

Sum of first $n$ terms: Let $S_n = a + (a+d)r + (a+2d)r^2 + \cdots + [a+(n-1)d]r^{n-1}$

Multiply by $r$: $rS_n = ar + (a+d)r^2 + \cdots + [a+(n-2)d]r^{n-1} + [a+(n-1)d]r^n$

Subtract: $S_n - rS_n = a + dr + dr^2 + \cdots + dr^{n-1} - [a+(n-1)d]r^n$

If $r \neq 1$: $S_n = \frac{a}{1-r} + \frac{dr(1-r^{n-1})}{(1-r)^2} - \frac{[a+(n-1)d]r^n}{1-r}$

Sum to infinity: For $|r| < 1$ and as $n \to \infty$:

$S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$

6.7 Solved Examples on Special Series

Example 1: Find $1^2 + 3^2 + 5^2 + \cdots$ to $n$ terms.

Solution:

The series is sum of squares of odd numbers.

$n$-th odd number = $2n-1$

We need: $\sum_{k=1}^{n} (2k-1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1)$

$= 4\sum_{k=1}^{n} k^2 - 4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1$

$= 4 \times \frac{n(n+1)(2n+1)}{6} - 4 \times \frac{n(n+1)}{2} + n$

$= \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n$

$= \frac{n}{3}[2(n+1)(2n+1) - 6(n+1) + 3]$

$= \frac{n}{3}[4n^2 + 6n + 2 - 6n - 6 + 3]$

$= \frac{n}{3}(4n^2 - 1) = \frac{n(4n^2 - 1)}{3}$

Example 2: Find sum to $n$ terms: $1 + 11 + 111 + 1111 + \cdots$

Solution:

Let $S_n = 1 + 11 + 111 + 1111 + \cdots$ to $n$ terms

Multiply by 9: $9S_n = 9 + 99 + 999 + 9999 + \cdots$

$= (10-1) + (100-1) + (1000-1) + (10000-1) + \cdots$

$= (10 + 10^2 + 10^3 + \cdots + 10^n) - n$

$= \frac{10(10^n - 1)}{10 - 1} - n = \frac{10(10^n - 1)}{9} - n$

So $S_n = \frac{1}{9}\left[\frac{10(10^n - 1)}{9} - n\right] = \frac{10(10^n - 1)}{81} - \frac{n}{9}$

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7. Convergence and Divergence of Infinite Series

7.1 Partial Sums

For an infinite series $\sum_{n=1}^{\infty} a_n$, define the partial sums:

$S_1 = a_1$

$S_2 = a_1 + a_2$

$S_3 = a_1 + a_2 + a_3$

$\vdots$

$S_n = a_1 + a_2 + \cdots + a_n = \sum_{k=1}^{n} a_k$

7.2 Convergence of Infinite Series

The infinite series $\sum_{n=1}^{\infty} a_n$ converges if the sequence of partial sums $\{S_n\}$ converges to a finite limit $S$.

That is, $\lim_{n \to \infty} S_n = S$ exists and is finite.

$S$ is called the sum of the infinite series.

If the limit does not exist or is infinite, the series diverges.

7.3 Important Tests for Convergence

a) Geometric Series Test

The geometric series $\sum_{n=0}^{\infty} ar^n$:

Converges if $|r| < 1$, and sum = $\frac{a}{1-r}$

Diverges if $|r| \geq 1$

b) $n$-th Term Test (Divergence Test)

If $\lim_{n \to \infty} a_n \neq 0$, then $\sum a_n$ diverges.

Note: If $\lim_{n \to \infty} a_n = 0$, the series may converge OR diverge.

c) p-Series Test

The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$:

Converges if $p > 1$

Diverges if $p \leq 1$

Special Cases:

Harmonic series $\sum \frac{1}{n}$ diverges ($p=1$)

$\sum \frac{1}{n^2}$ converges ($p=2>1$)

7.4 Solved Examples on Convergence

Example 1: Test convergence of $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$

Solution:

Using partial fractions: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

So $S_n = \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right)$

This is a telescoping series:

$S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$

$= 1 - \frac{1}{n+1}$

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1$

So series converges to 1.

Example 2: Test convergence of $\sum_{n=1}^{\infty} \frac{n}{2^n}$

Solution:

This is an AGP: $a_n = n \cdot \left(\frac{1}{2}\right)^n$

Let $S = \sum_{n=1}^{\infty} \frac{n}{2^n}$

Consider $S = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdots$

Multiply by $\frac{1}{2}$: $\frac{S}{2} = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \cdots$

Subtract: $S - \frac{S}{2} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$

$\frac{S}{2} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$

So $S = 2$

Thus series converges to 2.

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8. Arithmetic and Geometric Mean Properties

8.1 For Two Numbers

For two positive numbers $a$ and $b$:

Arithmetic Mean: $\text{AM} = \frac{a+b}{2}$

Geometric Mean: $\text{GM} = \sqrt{ab}$

Harmonic Mean: $\text{HM} = \frac{2ab}{a+b}$

Relationship: $\text{AM} \geq \text{GM} \geq \text{HM}$

Equality holds when $a = b$.

8.2 For $n$ Numbers

For $n$ positive numbers $a_1, a_2, \ldots, a_n$:

Arithmetic Mean: $\text{AM} = \frac{a_1 + a_2 + \cdots + a_n}{n}$

Geometric Mean: $\text{GM} = \sqrt[n]{a_1 \cdot a_2 \cdots a_n}$

Harmonic Mean: $\text{HM} = \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}$

Generalized Inequality: $\text{AM} \geq \text{GM} \geq \text{HM}$

8.3 Weighted Means

For numbers $a_1, a_2, \ldots, a_n$ with weights $w_1, w_2, \ldots, w_n$:

Weighted AM: $\frac{w_1 a_1 + w_2 a_2 + \cdots + w_n a_n}{w_1 + w_2 + \cdots + w_n}$

Weighted GM: $(a_1^{w_1} \cdot a_2^{w_2} \cdots a_n^{w_n})^{\frac{1}{w_1 + w_2 + \cdots + w_n}}$

8.4 Applications of AM-GM Inequality

Example: Prove that for positive $x$, $x + \frac{1}{x} \geq 2$

Solution:

By AM-GM inequality for $x$ and $\frac{1}{x}$:

$\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1$

So $x + \frac{1}{x} \geq 2$

Equality when $x = \frac{1}{x} \Rightarrow x^2 = 1 \Rightarrow x = 1$ (positive)

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9. Applications in Problem Solving

9.1 Word Problems on AP and GP

Example 1: A man saves Rs. 100 in January, Rs. 200 in February, Rs. 300 in March, and so on. How much will he save in December? What is his total savings for the year?

Solution:

Savings form an AP: $100, 200, 300, \ldots$

Here $a=100$, $d=100$

December is 12th month: $a_{12} = 100 + 11 \times 100 = 100 + 1100 = 1200$

Total for year: $S_{12} = \frac{12}{2}[2\times100 + 11\times100] = 6[200 + 1100] = 6 \times 1300 = 7800$

He saves Rs. 1200 in December and Rs. 7800 in total.

Example 2: The bacteria count in a culture doubles every hour. If initial count was 1000, what will be count after 6 hours? After how many hours will it reach 1,000,000?

Solution:

This is a GP with $a=1000$, $r=2$

After 6 hours: $a_7 = 1000 \times 2^6 = 1000 \times 64 = 64000$

Let after $n$ hours count reaches 1,000,000:

$1000 \times 2^{n-1} = 1000000$

$2^{n-1} = 1000$

Taking log: $(n-1)\log 2 = \log 1000 = 3$

$n-1 = \frac{3}{\log 2} \approx \frac{3}{0.3010} \approx 9.97$

$n \approx 10.97 \approx 11$ hours

9.2 Miscellaneous Applications

Example: If $a, b, c$ are in AP; $b, c, d$ are in GP; and $c, d, e$ are in HP, prove that $a, c, e$ are in GP.

Solution:

Given: $a, b, c$ are in AP $\Rightarrow 2b = a + c$ ...(1)

$b, c, d$ are in GP $\Rightarrow c^2 = bd$ ...(2)

$c, d, e$ are in HP $\Rightarrow d = \frac{2ce}{c+e}$ ...(3)

From (1): $b = \frac{a+c}{2}$

From (2): $d = \frac{c^2}{b} = \frac{c^2}{(a+c)/2} = \frac{2c^2}{a+c}$

From (3): $d = \frac{2ce}{c+e}$

Equate expressions for $d$:

$\frac{2c^2}{a+c} = \frac{2ce}{c+e}$

Cancel 2 and $c$ (assuming $c \neq 0$):

$\frac{c}{a+c} = \frac{e}{c+e}$

Cross multiply: $c(c+e) = e(a+c)$

$c^2 + ce = ae + ce$

$c^2 = ae$

Thus $a, c, e$ are in GP.

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10. Important Formulas Summary

10.1 Arithmetic Progression

$n$-th term: $a_n = a + (n-1)d$

Sum of first $n$ terms: $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + a_n)$

Arithmetic Mean of two numbers: $\text{AM} = \frac{a+b}{2}$

10.2 Geometric Progression

$n$-th term: $a_n = ar^{n-1}$

Sum of first $n$ terms: $S_n = \frac{a(1-r^n)}{1-r}$ for $r \neq 1$

Sum to infinity: $S_{\infty} = \frac{a}{1-r}$ for $|r| < 1$

Geometric Mean of two numbers: $\text{GM} = \sqrt{ab}$

10.3 Harmonic Progression

Harmonic Mean of two numbers: $\text{HM} = \frac{2ab}{a+b}$

Relation: For positive numbers, $\text{AM} \geq \text{GM} \geq \text{HM}$

10.4 Special Sums

Sum of first $n$ natural numbers: $\frac{n(n+1)}{2}$

Sum of squares: $\frac{n(n+1)(2n+1)}{6}$

Sum of cubes: $\left[\frac{n(n+1)}{2}\right]^2$

Sum of first $n$ odd numbers: $n^2$

Sum of first $n$ even numbers: $n(n+1)$

10.5 Selection of Terms

For AP:

Three terms: $a-d, a, a+d$

Four terms: $a-3d, a-d, a+d, a+3d$

For GP:

Three terms: $\frac{a}{r}, a, ar$

Four terms: $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$

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11. Exam Tips and Common Mistakes

11.1 Common Mistakes to Avoid

1. Confusing AP and GP formulas: Remember AP has addition, GP has multiplication

2. Incorrect $n$-th term: $a_n = a + (n-1)d$ for AP, $a_n = ar^{n-1}$ for GP

3. Wrong sum formula for GP: Use $S_n = \frac{a(1-r^n)}{1-r}$ when $|r| < 1$, or $S_n = \frac{a(r^n-1)}{r-1}$ when $|r| > 1$

4. Forgetting to check convergence for infinite GP: $|r| < 1$ required

5. Misapplying AM-GM inequality: Only valid for positive numbers

11.2 Problem-Solving Strategies

1. Identify the type: Check if sequence is AP, GP, HP, or something else

2. Write given information: Note first term, common difference/ratio, number of terms

3. Choose appropriate formula: Based on what's asked (term, sum, etc.)

4. Solve systematically: Show steps clearly

5. Verify answer: Check if reasonable (e.g., number of terms should be positive integer)

11.3 Quick Checks

1. For AP: Difference between consecutive terms is constant

2. For GP: Ratio between consecutive terms is constant

3. For HP: Reciprocals form AP

4. Sum formulas: Memorize the basic ones for AP and GP

5. AM-GM: Always $\text{AM} \geq \text{GM}$ for positive numbers

This comprehensive theory covers all aspects of sequences and series with detailed explanations and examples, providing complete preparation for the entrance examination.