Sequences, common differences, general terms, symmetric selection of numbers, and summation formulas.
1. Sequence
A sequence is an ordered arrangement of numbers formed according to a definite rule or pattern. The numbers occurring in a sequence are called its terms, denoted by $a_1, a_2, a_3, \dots, a_n$.
Finite vs Infinite Sequences: • Finite Sequence: A sequence with a countable number of terms, containing a definite last term (e.g., $2, 4, 6, 8, 10$).
• Infinite Sequence: A sequence that continues indefinitely without an end (e.g., $1, 4, 9, 16, \dots, n^2, \dots$).
2. Arithmetic Progression (AP)
An Arithmetic Progression (AP) is a sequence in which each term is obtained by adding a fixed number to the preceding term, except the first term.
First Term ($a$ or $a_1$): The initial term of the progression.
Common Difference ($d$): The constant difference between any consecutive pair of terms:
$$d = a_{k+1} - a_k \quad \text{for every positive integer } k$$
The common difference $d$ can be positive (increasing AP), negative (decreasing AP), or zero (constant AP).
Each subsequent term is generated by adding the common difference $d$.
3. Arithmetic Series
If $a_1, a_2, a_3, \dots, a_n$ is an arithmetic progression, then the indicated sum of its terms is called an arithmetic series:
Sequence vs Series: A sequence is an ordered set of elements separated by commas ($2, 5, 8, 11$), whereas a series is the algebraic sum of the terms ($2 + 5 + 8 + 11$).
4. General Term ($n^{\text{th}}$ Term) of an AP
Let the first term of an AP be $a$ and the common difference be $d$. By building the sequence term by term:
Derivation of General Term
$$\begin{aligned} a_1 &= a = a + (1 - 1)d \\ a_2 &= a_1 + d = a + d = a + (2 - 1)d \\ a_3 &= a_2 + d = (a + d) + d = a + 2d = a + (3 - 1)d \\ a_4 &= a_3 + d = (a + 2d) + d = a + 3d = a + (4 - 1)d \end{aligned}$$
Observing the pattern, the coefficient of $d$ is always one less than the term index $n$.
Therefore, the $n^{\text{th}}$ term (general term) $a_n$ is:
$$a_n = a + (n - 1)d$$
General Term Formula
$$a_n = a + (n - 1)d$$
5. $n^{\text{th}}$ Term from the End of an AP
Consider an AP having first term $a$, common difference $d$, and last term $l$ with $m$ total terms.
Traversing the AP backwards: first term becomes $l$ and common difference becomes $-d$.
Derivation
Method 1 (Reversing the AP): When viewed from the end to the beginning, the last term $l$ becomes the first term, and the common difference becomes $-d$:
$$a_n' = l + (n - 1)(-d) = l - (n - 1)d$$
Method 2 (From Beginning Index): In a finite AP with $m$ terms, the $n^{\text{th}}$ term from the end corresponds to the $(m - n + 1)^{\text{th}}$ term from the beginning:
$$a_{m-n+1} = a + [(m - n + 1) - 1]d = a + (m - n)d$$
Formula
$$a_n\text{ from end} = l - (n - 1)d$$
6. Arithmetic Mean (AM) Between Two Numbers
If three numbers $a, A, b$ are in arithmetic progression, then the middle number $A$ is called the Arithmetic Mean (AM) between $a$ and $b$.
Proof
Since $a, A, b$ are in AP, the common difference between consecutive terms must be equal:
$$A - a = b - A$$
$$A + A = a + b$$
$$2A = a + b \implies A = \frac{a + b}{2}$$
Arithmetic Mean Formula
$$A = \frac{a + b}{2}$$
7. Selection of 3, 4, and 5 Numbers in AP
When the sum of consecutive terms in an AP is given in a problem, choosing the terms symmetrically simplifies calculations because the common difference cancels out when summing:
3 Numbers in AP
Common difference is $d$. Summing the terms directly cancels $d$ to give $3a$.
$$a - d, \; a, \; a + d$$
$\text{Sum} = (a-d) + a + (a+d) = 3a$
4 Numbers in AP
Common difference is $2d$. Symmetrical around the center; summing cancels $d$.
$$a - 3d, \; a - d, \; a + d, \; a + 3d$$
$\text{Sum} = 4a \quad (\text{Common diff} = 2d)$
5 Numbers in AP
Common difference is $d$. Centered at $a$; summing yields $5a$.
$$a - 2d, \; a - d, \; a, \; a + d, \; a + 2d$$
$\text{Sum} = 5a \quad (\text{Common diff} = d)$
8. Sum of First $n$ Terms of an AP ($S_n$)
The sum of the first $n$ terms of an AP, denoted $S_n$, is derived by adding the series to itself in reverse order (Gauss's summation technique).
Derivation of Sum Formula
1. Write the sum of the first $n$ terms forward:
$$S_n = a + (a + d) + (a + 2d) + \dots + [a + (n - 2)d] + [a + (n - 1)d] \quad \text{--- (1)}$$
2. Write the series in reverse order:
$$S_n = [a + (n - 1)d] + [a + (n - 2)d] + \dots + (a + d) + a \quad \text{--- (2)}$$
Sum Formula (in terms of $a$, $d$, $n$)
$$S_n = \frac{n}{2} \big[ 2a + (n - 1)d \big]$$
9. Sum of an AP in Terms of the Last Term ($l$)
When the first term $a$, the total number of terms $n$, and the last term $l = a_n$ are known, the summation formula can be written in compact form.
Derivation
Splitting the $2a$ term inside the sum formula:
$$S_n = \frac{n}{2} \big[ 2a + (n - 1)d \big] = \frac{n}{2} \big[ a + \underbrace{a + (n - 1)d}_{l} \big]$$
Since the last term is $l = a + (n - 1)d$:
$$S_n = \frac{n}{2} (a + l)$$
Sum Formula (in terms of $a$ and $l$)
$$S_n = \frac{n}{2} (a + l)$$
Finding the $n^{\text{th}}$ term using sums: The $n^{\text{th}}$ term can always be computed by subtracting the sum of the first $(n-1)$ terms from the sum of the first $n$ terms:
$$a_n = S_n - S_{n-1} \quad (\text{for } n \ge 2), \quad \text{with } a_1 = S_1$$