7.3 Arithmetic Progressions (AP), General Term, Sum of n Terms & Sequence Modeling

Definition & Fundamental Parameters of an AP

An Arithmetic Progression (AP) is a sequence of real numbers in which each term after the first is obtained by adding a fixed constant, called the common difference ($d$), to the preceding term.

$$\text{General Sequence: } a, \; a+d, \; a+2d, \; a+3d, \; \dots, \; a+(n-1)d$$

• First Term ($a$): The initial value $T_1 = a$.
• Common Difference ($d$): The constant difference between any consecutive pair of terms:

$$d = T_n - T_{n-1} = T_2 - T_1 = T_3 - T_2$$

The common difference $d$ can be positive (increasing AP, e.g., $4, 7, 10, \dots$), negative (decreasing AP, e.g., $20, 16, 12, \dots$), or zero (constant sequence, e.g., $5, 5, 5, \dots$).

General $n$-th Term Formula

The $n$-th term from the beginning ($T_n$ or $a_n$) is given by:

$$T_n = a + (n - 1)d$$

$n$-th Term from the End of a Finite AP

If a finite AP contains last term $l$ and common difference $d$, counting backwards turns the sequence into an AP with first term $l$ and common difference $-d$:

$$T_n' = l - (n - 1)d$$

Total Number of Terms ($n$) in a Finite AP

If the first term is $a$, the last term is $l$, and the common difference is $d$:

$$l = a + (n - 1)d \implies n = \frac{l - a}{d} + 1$$

+-----------------------------------------------------------------------------------------+
|                           AP FUNDAMENTAL PARAMETERS OVERVIEW                            |
+-----------------------------------------------------------------------------------------+
| Starting Term:              a                                                           |
| Common Difference:          d = T_k - T_(k-1)                                           |
| n-th Term (from start):     T_n = a + (n - 1)d                                          |
| n-th Term (from end):       T_n' = l - (n - 1)d                                         |
| Term Count:                 n = [(l - a) / d] + 1                                       |
| Arithmetic Mean (2 terms):  AM = (a + b) / 2                                            |
+-----------------------------------------------------------------------------------------+

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Sum of the First $n$ Terms ($S_n$)

The sum of the first $n$ terms of an AP ($S_n = T_1 + T_2 + \dots + T_n$) is derived using Gauss's summation principle of writing the series forwards and backwards.

Derivation of Sum Formulas

$$S_n = a + (a + d) + (a + 2d) + \dots + (l - d) + l$$ $$S_n = l + (l - d) + (l - 2d) + \dots + (a + d) + a$$

Adding both expressions column-by-column yields $n$ identical pairs each summing to $(a + l)$:

$$2S_n = n(a + l) \implies S_n = \frac{n}{2}(a + l)$$

Substituting the last term $l = a + (n - 1)d$:

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

Recovering $T_n$ from Sum Function $S_n$

The $n$-th term is the marginal difference between consecutive cumulative sums:

$$T_n = S_n - S_{n-1} \quad (\text{for } n \ge 2), \quad \text{with } T_1 = S_1$$

The Quadratic Form Theorem for AP Sums

If the sum of $n$ terms of a sequence is represented as a quadratic polynomial in $n$ without a constant term:

$$S_n = An^2 + Bn$$

then the sequence is guaranteed to be an Arithmetic Progression with:

• Common Difference: $d = 2A$ (twice the coefficient of $n^2$)
• First Term: $a = S_1 = A(1)^2 + B(1) = A + B$
• General Term: $T_n = 2An + (B - A)$

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Arithmetic Mean (AM) & Insertion of Means

Single Arithmetic Mean between Two Numbers

If three numbers $a, M, b$ form an AP, then $M - a = b - M \implies 2M = a + b$. The middle term $M$ is the Arithmetic Mean (AM):

$$M = \frac{a + b}{2}$$

Inserting $n$ Arithmetic Means between $a$ and $b$

Let $A_1, A_2, A_3, \dots, A_n$ be $n$ numbers inserted between $a$ and $b$ such that the sequence:

$$a, \; A_1, \; A_2, \; A_3, \; \dots, \; A_n, \; b$$

forms an Arithmetic Progression. This progression contains a total of $N = n + 2$ terms.

1. The $(n+2)$-th term is $b$:

$$b = T_{n+2} = a + (n + 2 - 1)d = a + (n + 1)d$$

1. The common difference is therefore:

$$d = \frac{b - a}{n + 1}$$

1. The $k$-th inserted arithmetic mean $A_k$ is:

$$A_k = a + k \cdot d = a + k \left( \frac{b - a}{n + 1} \right)$$

1. Sum of all $n$ Inserted Means:

$$\sum_{k=1}^n A_k = n \times \left( \frac{a + b}{2} \right) = n \times (\text{AM of } a \text{ and } b)$$

+-----------------------------------------------------------------------------------------+
|                        INSERTION OF n ARITHMETIC MEANS                                  |
+-----------------------------------------------------------------------------------------+
| Given endpoints: a and b                                                                |
| Inserted Sequence: a, [A_1, A_2, A_3, ..., A_n], b   (Total: n + 2 terms)               |
|                                                                                         |
| Step 1: Compute Step Size:   d = (b - a) / (n + 1)                                      |
| Step 2: Compute k-th Mean:   A_k = a + k * d                                            |
| Step 3: Sum of all Means:    Sum = n * [(a + b) / 2]                                    |
+-----------------------------------------------------------------------------------------+

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Symmetric Term Selection Strategies

When a problem specifies properties regarding the sum and product of consecutive terms in an AP, choosing variables symmetrically around the center eliminates the common difference $d$ upon summation, leaving a single linear equation for $a$.

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In the comparison below, the finite GP sum formula requires $r\ne1$; for $r=1$ the sum is $na$. The geometric mean and AM–GM inequality use nonnegative endpoints $a,b$. An infinite GP with nonzero first term has sum $a/(1-r)$ only for $|r|<1$.

Comparison: Arithmetic Progression vs. Geometric Progression

While an AP grows or diminishes through constant additive increments, a Geometric Progression (GP) evolves via constant multiplicative factors.

+-----------------------------------------------------------------------------------------+
|                                 AP VS. GP ARCHITECTURE                                  |
+-----------------------------------------------------------------------------------------+
| Dimension:                 Arithmetic Progression (AP)    Geometric Progression (GP)    |
| Formation Rule:            Additive: T_n - T_(n-1) = d    Multiplicative: T_n/T_(n-1)=r |
| General Term:              T_n = a + (n - 1)d             T_n = a * r^(n - 1)           |
| Sum of n Terms:            S_n = (n/2)[2a + (n-1)d]       S_n = a(r^n - 1) / (r - 1)    |
| Infinite Series Sum:       Diverges (unless a=d=0)        S_inf = a / (1 - r)  (|r| < 1)|
| Mean Between a and b:      AM = (a + b) / 2               GM = sqrt(ab)                 |
| Mean Inequality:                               AM >= GM                                 |
+-----------------------------------------------------------------------------------------+

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Worked Examples

Worked Example 1: Sum of Decreasing AP & Double Answer Interpretation

Problem: How many terms of the arithmetic progression $17, 15, 13, \dots$ must be summed to yield a total sum of 72? Provide a rigorous physical and mathematical explanation for the occurrence of two distinct answers.

• Step 1: Identify parameters: First term $a = 17$, common difference $d = 15 - 17 = -2$, target sum $S_n = 72$.
• Step 2: Set up the sum equation:

$$S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]$$ $$72 = \frac{n}{2} \left[ 2(17) + (n - 1)(-2) \right]$$ $$72 = \frac{n}{2} \left[ 34 - 2n + 2 \right] = \frac{n}{2} [36 - 2n] = n(18 - n)$$

• Step 3: Solve the quadratic equation in $n$:

$$18n - n^2 = 72 \implies n^2 - 18n + 72 = 0$$

Factor the quadratic:

$$(n - 6)(n - 12) = 0 \implies n = 6 \quad \text{or} \quad n = 12$$

• Step 4: Explain the duality of solutions:
  • For $n = 6$: The first 6 terms are $17, 15, 13, 11, 9, 7$. Their sum is $17 + 15 + 13 + 11 + 9 + 7 = 72$.
  • Terms from $n = 7$ to $n = 12$ are: $5, 3, 1, -1, -3, -5$.
  • The sum of these additional six terms is:

$$5 + 3 + 1 + (-1) + (-3) + (-5) = 0$$

• Because the terms from the 7th to the 12th cancel out symmetrically around zero, the aggregate sum after 12 terms remains exactly 72. Both $n = 6$ and $n = 12$ are completely valid.

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Worked Example 2: Symmetric Term Selection & Cubic Products

Problem: The sum of three consecutive terms in an AP is 33, and the product of the first and the third term is 105. Find the three numbers.

• Step 1: Choose symmetric variables: Let the three terms be $a - d, \; a, \; a + d$.
• Step 2: Utilize the sum condition:

$$(a - d) + a + (a + d) = 33 \implies 3a = 33 \implies a = 11$$

• Step 3: Utilize the product condition:

$$(a - d)(a + d) = 105 \implies a^2 - d^2 = 105$$

Substitute $a = 11$:

$$11^2 - d^2 = 105 \implies 121 - d^2 = 105$$ $$d^2 = 121 - 105 = 16 \implies d = \pm 4$$

• Step 4: Write down the resulting terms:
  • If $d = +4$: Terms are $11 - 4, \; 11, \; 11 + 4 \implies 7, \; 11, \; 15$.
  • If $d = -4$: Terms are $11 - (-4), \; 11, \; 11 + (-4) \implies 15, \; 11, \; 7$.
• Conclusion: In either progression, the three terms are 7, 11, and 15.

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Worked Example 3: Railway Concrete Sleeper Deployment Modeling

Problem: A railway track maintenance gang is laying pre-stressed concrete sleepers along a new yard siding. The layout is divided into consecutive sections. The first section receives 24 sleepers, and each subsequent section receives 4 more sleepers than the previous one. If a total of 780 sleepers are laid across the entire siding, determine: (a) The total number of sections laid, and (b) The number of sleepers laid in the final section.

• Step 1: Identify progression parameters: First term $a = 24$, common difference $d = 4$, aggregate sum $S_n = 780$.
• Step 2: Formulate the sum equation:

$$S_n = \frac{n}{2} [2a + (n - 1)d] = 780$$ $$\frac{n}{2} [2(24) + (n - 1)(4)] = 780$$ $$\frac{n}{2} [48 + 4n - 4] = \frac{n}{2} [4n + 44] = n(2n + 22) = 2n^2 + 22n = 780$$

• Step 3: Solve the quadratic equation: Divide by 2:

$$n^2 + 11n - 390 = 0$$

Factor $390$: We seek two numbers with product $-390$ and difference $11$. Factors are $26$ and $15$ ($26 \times 15 = 390$, $26 - 15 = 11$):

$$(n + 26)(n - 15) = 0$$

Since the number of sections $n$ must be a positive integer, $n = 15$.

• Step 4: Compute sleepers in the final (15th) section:

$$T_{15} = a + (15 - 1)d = 24 + 14(4) = 24 + 56 = 80 \text{ sleepers}$$

• Conclusion: There are 15 sections, and the final section contains 80 sleepers.

The term-count formula dividing by $d$ assumes $d\ne0$. For $d=0$, every term equals $a$ and the endpoints alone do not determine the number of terms.