PA and PG: summary, formulas and exercises

THE arithmetic progression - PA is a sequence of values ​​that has a constant difference between consecutive numbers.

THE geometric progression - PG presents numbers with the same quotient when dividing two consecutive terms.

While in the arithmetic progression the terms are obtained by adding the difference common to the predecessor, the terms of a geometric progressions are found by multiplying the ratio by the last number in the sequence, thus obtaining the term successor.

Below is a summary of the two types of progressions.

Arithmetic Progression (AP)

An arithmetic progression is a sequence formed by terms that differ from each other by a constant value, which is called ratio, calculated by:

Where,

r is the reason for the BP;
The₂ is the second term;
The₁ is the first term.

Therefore, the terms of an arithmetic progression can be written as follows:

Note that in a PA of no terms the formula of the general term (the_no) of the sequence is:

The_no = the₁ + (n – 1) r

Some particular cases are: a 3-term AP is represented by (x - r, x, x + r) and a 5-term AP has its components represented by (x - 2r, x - r, x, x + r, x + 2r).

Types of PA

According to the ratio value, arithmetic progressions are classified into 3 types:

1. Constant: when the ratio is equal to zero and the BP terms are equal.

Example: PA = (2, 2, 2, 2, 2, ...), where r = 0

2. Growing: when the ratio is greater than zero and a term from the second is greater than the previous one;

Example: PA = (2, 4, 6, 8, 10, ...), where r = 2

3. descending: when the ratio is less than zero and a term from the second is less than the previous one.

Example: PA = (4, 2, 0, - 2, - 4, ...), where r = - 2

Arithmetic progressions can still be classified into finite, when they have a certain number of terms, and infinite, that is, with infinite terms.

Sum of terms of a PA

The sum of the terms of an arithmetic progression is calculated by the formula:

Where, no is the number of terms in the sequence, The₁ is the first term and The_no is the nth term. The formula is useful for solving questions where the first and last term is given.

When a problem has the first term and the BP reason, you can use the formula:

These two formulas are used to add the terms of a finite BP.

Average term of the PA

To determine the mean or central term of an BP with an odd number of terms we calculate the arithmetic mean with the first and last term (a₁ and the_no):

The mean term between three consecutive numbers of a PA corresponds to the arithmetic mean of the predecessor and successor.

Solved example

Given the PA (2, 4, 6, 8, 10, 12, 14) determine the ratio, the mean term and the sum of the terms.

1. PA reason

2. medium term

3. sum of terms

Learn more about arithmetic progression.

Geometric progression (PG)

A geometric progression is formed when a sequence has a multiplier factor resulting from dividing two consecutive terms, called a common ratio, which is calculated by:

Where,

what is the reason for PG;
The₂ is the second term;
The₁ is the first term.

A geometric progression of no terms can be represented as follows:

Being The₁ the first term, the general term of PG is calculated by The₁.q^(no-1).

PG Types

According to the value of the ratio (q), we can classify the Geometric Progressions into 4 types:

1. Growing: the ratio is always positive (q > 0) and the terms are increasing;

Example: PG: (3, 9, 27, 81, ...), where q = 3.

2. descending: the ratio is always positive (q > 0), non-zero (0), and the terms are decreasing;

Example: PG: (-3, -9, -27, -81, ...), where q = 3

3. oscillating: the reason is negative (q

Example: PG: (3, -6, 12, -24, 48, -96, …), where q = - 2

4. Constant: the ratio is always equal to 1 and the terms have the same value.

Example: PG: (3, 3, 3, 3, 3, 3, 3, ...), where q = 1

Sum of terms of a PG

The sum of terms of a geometric progression is calculated by the formula:

Being The₁ the first term, what the common reason and no the number of terms.

If the PG ratio is less than 1, then we will use the following formula to determine the sum of terms.

These formulas are used for a finite PG. If the requested sum is an infinite PG, the formula used is:

Average term of PG

To determine the mean or central term of a PG with an odd number of terms we calculate the geometric mean with the first and last term (a₁ and the_no):

Solved example

Given PG (1, 3, 9, 27, and 81) determine the ratio, the average term, and the sum of the terms.

1. PG reason

2. medium term

3. sum of terms

Learn more about geometric progression.

Summary of PA and PG formulas

arithmetic progression	Geometric progression
Reason
general term
medium term
finite sum
infinite sum

Learn more about number sequences.

Exercises on PA and PG

question 1

What is the 16th term of the sequence that starts with the number 3 and has an BP ratio equal to 4?

a) 36
b) 52
c) 44
d) 63

See Answer

Correct alternative: d) 63.

Since the ratio of a PA is constant, we can find the second term in the sequence by adding the ratio to the first number.

The₂ = the₁ + r

The₂ = 3 + 4

The₂ = 7

Therefore, we can say that this sequence is formed by (3, 7, 11, 15, 19, 23, …)

The 16th term can be calculated with the general term formula.

The_no = the₁ + (n - 1). r

The₁₆ = 3 + (16 – 1). 4

The₁₆ = 3 + 15.4

The₁₆ = 3 + 60

The₁₆ = 63

Therefore, the answer to the question is 63.

question 2

What is the ratio of a six-term AP whose sum of the first three numbers in the sequence equals 12 and the last two equals –34?

a) 7
b) - 6
c) – 5
d) 5

See Answer

Correct alternative: b) – 6.

The general formula for the terms of an arithmetic progression is₁, (a₁ + r), (a₁ + 2r),..., {a₁ + (n-1) r}. Therefore, the sum of the first three terms can be written as follows:

The₁ + (the₁ + r) + (a₁ + 2r) = 12
3rd₁ + 3r = 12
3rd₁ = 12 - 3r
The₁ = (12 - 3r)/3
The₁ = 4 - r

And the sum of the last two terms is:

(The₁ + 4r) + (a₁ + 5r) = – 34
2nd₁ + 9r = – 34

Now we replace the₁ by 4 – r.

2(4 – r) + 9r = – 34
8 – 2r + 9r = – 34
7r = – 34 – 8
7r = – 42
r = – 42/7
r = – 6

Therefore, the PG ratio is - 6.

question 3

If the third term of a GP is 28 and the fourth term is 56 what are the first 5 terms of this geometric progression?

a) 6, 12, 28, 56, 104
b) 7, 18, 28, 56, 92
c) 5, 9, 28, 56, 119
d) 7, 14, 28, 56, 112

See Answer

Correct alternative: d) 7, 14, 28, 56, 112

First, we must calculate the ratio of this PG. For this, we will use the formula:

The₄ = the₃. what
56 = 28. what
56 / 28 = q
q = 2

Now we calculate the first 5 terms. We'll start with the₁ using the formula of the general term.

The_no = the₁. what^(n-1)
The₃ = the₁ . what^(3-1)
28 = the₁. 2²
The₁ = 28/ 4 = 7

The remaining terms can be calculated by multiplying the antecedent term by the ratio.

The₂ = the₁.q
The₂ = 7. 2
The₂ = 14

The₅ = the₄. what
The₅ = 56. 2
The₅ = 112

Therefore, the first 5 terms of PG are:

1st term: 7
2nd term: 14
3rd term: 28
4th term: 56
5th term: 112

See also other exercises to keep practicing:

• Exercises on Arithmetic Progression
• Exercises on Geometric Progression