[Free Cheatsheet] Sequence and Series - A.P, G.P, A.G.P, & H.P

Arithmetic Progression (AP)

An arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This constant is called the common difference ($d$).

General Term (nth term) of AP:
$a_n = a + (n - 1)d$
Where:
$a$ = first term,
$d$ = common difference,
$n$ = term number.
Example: For an AP with $a = 2$, $d = 3$, find the 5th term.
Solution: $a_5 = 2 + (5 - 1) \cdot 3 = 2 + 12 = 14$.

Sum of First n Terms of AP:
$S_n = \frac{n}{2} [2a + (n - 1)d]$
Alternatively: $S_n = \frac{n}{2} (a + l)$.
Where:
$a$ = first term,
$l$ = last term,
$n$ = number of terms.
Example: Find the sum of the first 10 terms of the AP 2, 5, 8, ....
Solution: $a = 2, d = 3, n = 10$.
$S_{10} = \frac{10}{2} [2 \cdot 2 + (10 - 1) \cdot 3] = 5[4 + 27] = 5 \cdot 31 = 155$.

Common Difference:
$d = a_2 - a_1$
Where:
$a_1$ = first term,
$a_2$ = second term.
Example: If the first two terms of an AP are 7 and 12, find $d$.
Solution: $d = 12 - 7 = 5$.

Arithmetic Mean (AM):
$AM = \frac{a + b}{2}$
Where: $a$ and $b$ are two numbers.
Example: Find the arithmetic mean of 4 and 10.
Solution: $AM = \frac{4 + 10}{2} = \frac{14}{2} = 7$.

Properties:
- If three numbers $a, b, c$ are in AP, then $2b = a + c$.
Example: Show that 3, 7, 11 are in AP.
Solution: $2 \cdot 7 = 3 + 11 \Rightarrow 14 = 14$ (True).
- Sum of an AP is symmetric around the middle term when the number of terms is odd.

Question 1:
Find the sum of the first 20 terms of an AP where the first term is 5 and the 20th term is 62.
Solution:
$S_n = \frac{n}{2} (a + l)$.
$S_{20} = \frac{20}{2} (5 + 62) = 10 \cdot 67 = 670$.

Question 2:
If the 5th term of an AP is 20 and the 10th term is 45, find the first term and the common difference.
Solution:
General term: $a_n = a + (n - 1)d$.
For the 5th term: $a_5 = a + 4d = 20$.
For the 10th term: $a_{10} = a + 9d = 45$.
Solve the system of equations:
$a + 4d = 20$
$a + 9d = 45$
Subtract: $5d = 25 \Rightarrow d = 5$
Substitute: $a + 4 \cdot 5 = 20 \Rightarrow a = 0$.
First term: $a = 0$, common difference: $d = 5$.

Additional Formula:
- Number of terms: $n = \frac{l - a}{d} + 1$
Where: $l$ = last term.
Example: Find the number of terms in the AP 3, 7, 11, ..., 47.
Solution: $n = \frac{47 - 3}{4} + 1 = \frac{44}{4} + 1 = 11 + 1 = 12$.
There are 12 terms in the AP.

Geometric Progression (GP)

A geometric progression (GP) is a sequence of numbers where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio ($r$).

General Term (nth term) of GP:
$a_n = ar^{n-1}$
Where:
$a$ = first term,
$r$ = common ratio,
$n$ = term number.
Example: For a GP with $a = 2$ and $r = 3$, find the 4th term.
Solution: $a_4 = 2 \cdot 3^{4-1} = 2 \cdot 27 = 54$.

Sum of First n Terms of GP:
$S_n = a \frac{r^n - 1}{r - 1}$
For $r \neq 1$.
Example: Find the sum of the first 4 terms of the GP $2, 6, 18, 54...$.
Solution: $a = 2, r = 3, n = 4$.
$S_4 = 2 \frac{3^4 - 1}{3 - 1} = 2 \frac{81 - 1}{2} = 2 \cdot 40 = 80$.

Sum to Infinity of GP (when $|r| < 1$):
$S_\infty = \frac{a}{1 - r}$
Example: For a GP with $a = 5$ and $r = 0.5$, find the sum to infinity.
Solution: $S_\infty = \frac{5}{1 - 0.5} = \frac{5}{0.5} = 10$.

Common Ratio:
$r = \frac{a_2}{a_1}$
Where $a_2$ = second term and $a_1$ = first term.
Example: If the first two terms of a GP are $4$ and $12$, find $r$.
Solution: $r = \frac{12}{4} = 3$.

Geometric Mean (GM):
$GM = \sqrt{ab}$
For two numbers $a$ and $b$.
Example: Find the geometric mean of $4$ and $16$.
Solution: $GM = \sqrt{4 \cdot 16} = \sqrt{64} = 8$.

Properties:
- If three numbers $a, b, c$ are in GP, then $b^2 = ac$.
Example: Show that $2, 6, 18$ are in GP.
Solution: $6^2 = 2 \cdot 18 \Rightarrow 36 = 36$ (True).

Additional Formula:
- Number of terms: $n = \frac{\log(l/a)}{\log(r)} + 1$
Where $l$ = last term.
Example: Find the number of terms in the GP $2, 6, 18, ..., 486$.
Solution: $n = \frac{\log(486/2)}{\log(3)} + 1 = \frac{\log(243)}{\log(3)} + 1 = \frac{5}{1} + 1 = 6$.
There are 6 terms in the GP.

Arithmetic-Geometric Progression (AGP)

An Arithmetic-Geometric Progression (AGP) is a sequence where each term is the product of the corresponding terms of an arithmetic progression (AP) and a geometric progression (GP).
For example, if $\{a_1, a_2, a_3, \ldots\}$ is an AP and $\{b_1, b_2, b_3, \ldots\}$ is a GP, then the sequence $\{a_1b_1, a_2b_2, a_3b_3, \ldots\}$ is an AGP.

General Term (nth term) of AGP:
If the nth term of the AP is $a_n = a + (n - 1)d$ and the nth term of the GP is $b_n = br^{n-1}$, then the nth term of the AGP is given by:
$T_n = a_n \cdot b_n = \left(a + (n - 1)d\right) \cdot br^{n-1}$.

Example 1:
Given an AP with $a = 2$, $d = 3$, and a GP with $b = 4$, $r = 2$, find the 4th term of the AGP.
Solution: $T_4 = \left(2 + (4 - 1) \cdot 3\right) \cdot 4 \cdot 2^{4-1} = \left(2 + 9\right) \cdot 4 \cdot 8 = 11 \cdot 32 = 352$.

Sum of First n Terms of AGP:
The sum of the first n terms of an AGP is given by:
$S_n = \sum_{k=1}^n \left[a + (k - 1)d\right] \cdot br^{k-1}$.
To simplify, this can be split into two parts:
$S_n = b \cdot \sum_{k=1}^n ar^{k-1} + b \cdot \sum_{k=1}^n (k - 1)dr^{k-1}$.
These two sums are calculated separately:

• The first term, $\sum_{k=1}^n ar^{k-1}$, is a standard geometric progression with sum:
  $S_1 = a \cdot \frac{1 - r^n}{1 - r}$ for $r \neq 1$.
• The second term, $\sum_{k=1}^n (k - 1)dr^{k-1}$, involves advanced summation techniques.

Special Case:
When the common ratio $r$ of the GP satisfies $|r| < 1$, the sum to infinity of the AGP can be calculated.
The sum to infinity is given by:
$S_\infty = \frac{ab}{1 - r} + \frac{bdr}{(1 - r)^2}$.

Example 2:
Consider an AGP where the AP is $2, 4, 6, \ldots$ (with $a = 2, d = 2$) and the GP is $1, 0.5, 0.25, \ldots$ (with $b = 1, r = 0.5$). Find the sum to infinity of the AGP.
Solution: $S_\infty = \frac{2 \cdot 1}{1 - 0.5} + \frac{1 \cdot 2 \cdot 0.5}{(1 - 0.5)^2} = \frac{2}{0.5} + \frac{1}{0.25} = 4 + 4 = 8$.

Example 3:
Find the sum of the first 5 terms of an AGP where the AP is $3, 6, 9, \ldots$ (with $a = 3, d = 3$) and the GP is $2, 4, 8, \ldots$ (with $b = 2, r = 2$).
Solution: Each term is calculated as:
$T_1 = 3 \cdot 2 = 6$, $T_2 = 6 \cdot 4 = 24$, $T_3 = 9 \cdot 8 = 72$, $T_4 = 12 \cdot 16 = 192$, $T_5 = 15 \cdot 32 = 480$.
$S_5 = 6 + 24 + 72 + 192 + 480 = 774$.

Properties of AGP:
- The nth term of an AGP combines the linear growth of an AP and the exponential growth/decay of a GP.
- The sum of an AGP often requires separating into two summable series (AP and GP contributions).

Harmonic Progression (HP)

A sequence is said to be in harmonic progression (HP) if the reciprocals of its terms form an arithmetic progression (AP).
For example, if $\{a_1, a_2, a_3, \ldots\}$ is in HP, then $\left\{\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots\right\}$ is in AP.

General Term (nth term) of HP:
If the terms of HP are $a, b, c, \ldots$, the nth term of the HP can be derived as follows:
$\frac{1}{a_n} = \frac{1}{a} + (n - 1)d'$
Where:
$a$ = first term of HP,
$d'$ = common difference of the corresponding AP of reciprocals.
Example: If the first term of HP is \(4\) and the common difference of its reciprocals AP is \(0.5\), find the 3rd term.
Solution: $\frac{1}{a_3} = \frac{1}{4} + (3 - 1) \cdot 0.5 = 0.25 + 1 = 1.25$
$a_3 = \frac{1}{1.25} = 0.8$.

nth Term of HP:
The nth term of an HP can also be expressed as:
$a_n = \frac{1}{\frac{1}{a} + (n - 1)d'}$
Where:
$\frac{1}{a}$ = first term of the corresponding AP of reciprocals,
$d'$ = common difference of the reciprocals AP.
Example: For the HP \(4, 3, 2.4, \ldots\), find the 4th term.
Solution: The corresponding AP of reciprocals is \(0.25, 0.333..., 0.416...\).
$\frac{1}{a_4} = \frac{1}{4} + (4 - 1)d' = 0.25 + 3 \cdot 0.0833 \approx 0.5$
$a_4 = \frac{1}{0.5} = 2$.

Harmonic Mean (HM):
The harmonic mean of two numbers $a$ and $b$ is given by:
$HM = \frac{2ab}{a + b}$
Example: Find the harmonic mean of \(4\) and \(6\).
Solution: $HM = \frac{2 \cdot 4 \cdot 6}{4 + 6} = \frac{48}{10} = 4.8$.

Properties of HP:
- If three terms $a, b, c$ are in HP, then their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP.
- The harmonic mean of two numbers is always less than or equal to their geometric mean.
Example: Show that \(6, 12, 24\) are in HP.
Solution: The reciprocals are \(0.166..., 0.083..., 0.0416...\), which form an AP with a common difference of \(-0.0416...\).

Decimal Series and Their Summation

A decimal series is a sequence of numbers where each term is a decimal or fraction, often forming a recognizable pattern. Decimal series are frequently encountered in mathematics, finance, and scientific calculations. Understanding their summation involves identifying the series' structure and applying appropriate formulas or methods.

General Form of a Decimal Series:
A decimal series can often be written in the form:
$S = a + \frac{b}{10} + \frac{c}{100} + \frac{d}{1000} + \ldots$.
Alternatively, it may represent a geometric series if the terms follow a constant ratio.

Summation of Finite Decimal Series:
For a finite series, summation is performed by adding all terms explicitly:
$S = a_1 + a_2 + \ldots + a_n$.
Example: Find the sum of the series $1.2 + 0.34 + 0.056$.
Solution: $S = 1.2 + 0.34 + 0.056 = 1.596$.

Summation of Infinite Decimal Series:
If the series is infinite and follows a geometric progression, the sum can be computed using:
$S_\infty = \frac{a}{1 - r}$.
Where:
$a$ = first term,
$r$ = common ratio (with $|r| < 1$).
Example: Find the sum of the infinite series $0.5 + 0.05 + 0.005 + \ldots$.
Solution: $a = 0.5, r = 0.1$.
$S_\infty = \frac{0.5}{1 - 0.1} = \frac{0.5}{0.9} = 0.555...$.

Periodic Decimal Series:
A periodic decimal series repeats in a pattern, such as $0.333\ldots$ or $0.142857142857\ldots$.
Such series can be expressed as fractions:
$0.\overline{3} = \frac{1}{3}$ and $0.\overline{142857} = \frac{1}{7}$.
Example: Convert $0.\overline{6}$ to a fraction.
Solution: Let $x = 0.\overline{6}$. Multiply by 10:
$10x = 6.\overline{6}$.
Subtract: $10x - x = 6 \Rightarrow 9x = 6 \Rightarrow x = \frac{2}{3}$.

Properties of Decimal Series:
- Decimal series can often be simplified into a fractional or geometric form.
- The sum of a periodic decimal series converges to a rational number.
- The representation of a decimal as a fraction depends on its periodicity or termination.

Application Example:
A machine reduces errors by 10% each cycle, starting with an error rate of 1. Calculate the total error rate after infinite cycles.
Solution: The error series is $1 + 0.1 + 0.01 + 0.001 + \ldots$.
This is a GP with $a = 1$ and $r = 0.1$.
$S_\infty = \frac{1}{1 - 0.1} = \frac{1}{0.9} = 1.111...$.

Question:
Find the sum of the series $0.25 + 0.025 + 0.0025 + \ldots$.
Solution:
The series is a GP with $a = 0.25$ and $r = 0.1$.
$S_\infty = \frac{0.25}{1 - 0.1} = \frac{0.25}{0.9} = 0.277...$.

Question:
Find the sum up to $n$ terms of the series:
$0.3 + 0.33 + 0.333 + \dots$.

Solution:
Each term in the series can be written as:
$a_k = \frac{10^k - 1}{10^k}$.
The sum of the series up to $n$ terms is:
$S_n = \sum_{k=1}^n \frac{10^k - 1}{10^k}$.
Breaking the summation into two parts:
$S_n = \sum_{k=1}^n 1 - \sum_{k=1}^n \frac{1}{10^k}$.
The first sum:
$\sum_{k=1}^n 1 = n$.
The second sum is a geometric series with first term $a = 0.1$ and common ratio $r = 0.1$:
$\sum_{k=1}^n \frac{1}{10^k} = \frac{0.1(1 - (0.1)^n)}{1 - 0.1}$.
Simplifying:
$\sum_{k=1}^n \frac{1}{10^k} = 0.111\ldots(1 - 0.1^n)$.
Therefore, the sum is:
$S_n = n - 0.111\ldots(1 - 0.1^n)$.
Example: For $n = 3$, the series is:
$S_3 = 3 - 0.111\ldots(1 - 0.001) = 3 - 0.111\ldots(0.999) = 2.888\ldots$.