Tenth Grade Mathematics: Arithmetic Progressions: General Term (n-th Term)

Among the vast diversity of sequences, the simplest and most widespread pattern is one where each consecutive term increases or decreases by a constant numerical difference. Such a sequence is called an Arithmetic Progression, commonly abbreviated as AP. From calculating uniform financial depreciation to predicting the height of a tiered staircase, arithmetic progressions provide a direct linear framework for modeling uniform additive growth.

Definition of an Arithmetic Progression

An Arithmetic Progression (AP) is a sequence of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term.

Formal Definition:
A sequence a1, a2, a3, ..., an is an AP if:
                     a_(k+1) - a_k = d
for all positive integers k, where d is a constant real number.

The first term of an AP is traditionally denoted by the letter a, and the constant difference d is called the common difference.

Behavior of the Common Difference d:
1. If d > 0, the AP is strictly increasing: (3, 7, 11, 15, ...) [d = +4]
2. If d < 0, the AP is strictly decreasing: (20, 15, 10, 5, ...) [d = -5]
3. If d = 0, the AP is constant: (4, 4, 4, 4, ...) [d = 0]

Notice that the common difference can be positive, negative, or zero, and it can be an integer, fraction, or irrational number.

Derivation of the General Term (n-th Term)

Let the first term of an AP be a and the common difference be d. We write out the terms sequentially:

Term 1: a1 = a = a + (1 - 1)*d
Term 2: a2 = a1 + d = a + d = a + (2 - 1)*d
Term 3: a3 = a2 + d = (a + d) + d = a + 2*d = a + (3 - 1)*d
Term 4: a4 = a3 + d = (a + 2*d) + d = a + 3*d = a + (4 - 1)*d
...

Observing the clear inductive pattern, the coefficient of d is always strictly one less than the term index n:

Theorem: The n-th term an of an Arithmetic Progression with first term a
and common difference d is given by:
                          an = a + (n - 1) * d

If an AP has a finite number of terms m, then the last term is often denoted by l, where l = a + (m - 1)*d.

The Linear Structure of an AP:
an = a + n*d - d = (d)*n + (a - d)
Notice that an is a linear polynomial in n, where the coefficient of n is
identically the common difference d!

Calculating Terms from the End of an AP

Frequently, problems ask for the n-th term measured from the end of a finite AP having m terms.

Formula for n-th Term from the End:
Method 1: An AP with last term l and common difference d becomes an AP with
          first term l and common difference -d when viewed in reverse!
          an (from end) = l - (n - 1) * d

Method 2: The n-th term from the end corresponds to the:
          (m - n + 1)-th term from the beginning.

Example: Find the 10th term from the end of the AP: 3, 8, 13, ..., 253.
Here last term l = 253, and common difference d = 8 - 3 = 5.
Using Method 1:
10th term from end = l - (10 - 1)*d = 253 - 9*(5) = 253 - 45 = 208.

Determining whether a Number Belongs to an AP

To determine whether a specific number K belongs to a given AP:

Set an = K: a + (n - 1)*d = K. Solve the equation for the index n: n = (K - a)/d + 1. Examine the value of n: If n is a positive integer (n in {1, 2, 3, ...}), then K is indeed the n-th term of the AP. If n is a fraction, negative, or not an integer, then K does not belong to the AP.

Chapter Exercises

Exercise 1: Find the 30th term of the AP: 10, 7, 4, ... and find the 11th term of the AP: -3, -1/2, 2, ...

Exercise 2: Which term of the AP: 3, 8, 13, 18, ... is 78? Also check whether -150 is a term of the AP: 11, 8, 5, 2, ...

Exercise 3: An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.

Solutions and Explanations

Solution 1: For the first AP (10, 7, 4, ...), first term a = 10, common difference d = 7 - 10 = -3. The 30th term is a30 = a + (30 - 1)d = 10 + 29(-3) = 10 - 87 = -77. For the second AP (-3, -1/2, 2, ...), first term a = -3, common difference d = -1/2 - (-3) = -1/2 + 3 = 5/2. The 11th term is a11 = a + (11 - 1)d = -3 + 10(5/2) = -3 + 25 = 22.

Solution 2: For the AP 3, 8, 13, 18, ..., first term a = 3 and common difference d = 8 - 3 = 5. To find which term is 78, set an = 78: 3 + (n - 1)5 = 78 ==> (n - 1)5 = 75 ==> n - 1 = 15 ==> n = 16. Thus, 78 is the 16th term of the AP. Now check if -150 is in 11, 8, 5, 2, ...: here a = 11, d = 8 - 11 = -3. Set an = -150: 11 + (n - 1)(-3) = -150 ==> (n - 1)(-3) = -161 ==> n - 1 = 161/3 ==> n = 161/3 + 1 = 164/3 = 54.666... Because n is not a natural number (it is a fraction), -150 cannot be a term of this AP.

Solution 3: We are given total terms m = 50. The 3rd term is a3 = a + 2d = 12 (Equation 1). The last term (50th term) is a50 = a + 49d = 106 (Equation 2). Subtract Equation 1 from Equation 2: (a + 49d) - (a + 2d) = 106 - 12 ==> 47d = 94 ==> d = 94/47 = 2. Substitute d = 2 into Equation 1: a + 2(2) = 12 ==> a + 4 = 12 ==> a = 8. Now calculate the 29th term: a29 = a + 28d = 8 + 28(2) = 8 + 56 = 64. Thus, the 29th term is 64.