An arithmetic progression is a sequence where each subsequent term is obtained by adding a constant number \(d\) (the common difference) to the previous term. Any term can be found using the formula \(a_n = a_1 + (n-1)d\), and the sum of the first \(n\) terms using the formula \(S_n = \dfrac{a_1 + a_n}{2}\cdot n\). This page covers the definition and common difference, both sum formulas, the characteristic property, checking if a number belongs to the progression, its difference from a geometric progression, typical OGE problems, and an interactive trainer for self-testing.
What is an Arithmetic Progression
An arithmetic progression is a numerical sequence where each subsequent term is obtained by adding the same number to the previous term. This number is called the common difference of the progression and is denoted by the letter \(d\).
Definition in one formula: a sequence \(a_1, a_2, a_3, \dots\) is called an arithmetic progression if for any index \(n\), the equality
holds.
Let's take the sequence 3, 7, 11, 15, 19, … The first term here is \(a_1 = 3\), and the common difference is \(d = 4\): each subsequent number is exactly 4 greater than the previous one. Therefore, the next term is 23, and if we were to continue backwards, the number before three would be minus one.
The entire progression is defined by just two numbers — the first term \(a_1\) and the common difference \(d\). Knowing this pair, you can determine any term, calculate the sum of any number of initial terms, and check if a given number appears in the sequence. The progression is denoted by the symbol \((a_n)\), and its terms by \(a_1\), \(a_2\), \(a_3\), and so on, where the subscript always indicates the index, not the value.
Common difference of an arithmetic progression: how to find d
The common difference is always the next term minus the previous term:
The order of subtraction is important: if you subtract in the reverse order, the sign of the difference will be opposite. For the sequence 20, 17, 14, 11, 8, the common difference is \(17 - 20 = -3\), not \(+3\).
The sign of \(d\) completely determines the behavior of the progression:
- \(d > 0\) — the progression is increasing, terms get larger;
- \(d < 0\) — the progression is decreasing, terms get smaller;
- \(d = 0\) — all terms are the same, e.g., 6, 6, 6, 6, … This is also an arithmetic progression, just a constant one.
If two terms with different indices are known, the common difference is found by the formula
Example. Suppose \(a_4 = 30\) and \(a_{10} = 6\). Then \(d = \dfrac{6 - 30}{10 - 4} = \dfrac{-24}{6} = -4\), and the first term is \(a_1 = a_4 - 3d = 30 + 12 = 42\). Let's check the beginning of the progression: 42, 38, 34, 30 — the fourth term is indeed 30.
The common difference is not necessarily an integer: for the sequence 1; 1.5; 2; 2.5, the common difference is 0.5, and it is a valid arithmetic progression.
Formula for the n-th term of an arithmetic progression
The main formula of the topic allows you to find any term without listing all the preceding ones:
Why \(n - 1\)? To get from the first term to the \(n\)-th term, the common difference is added not \(n\) times, but one less time: to reach the second term takes one step, to the third takes two, to the twentieth takes nineteen.
Example. For the progression 3, 7, 11, 15, 19, … we have \(a_1 = 3\) and \(d = 4\), so
This compact form is very convenient: it's immediately clear that \(a_{20} = 4\cdot 20 - 1 = 79\), and \(a_{100} = 399\). It also hints at the common difference: for any arithmetic progression, the formula for the \(n\)-th term is linear, of the form \(a_n = dn + c\), and the coefficient of \(n\) is the common difference.
Sometimes the first term is unknown, but another term is given. In this case, the same idea works in a general form: \(a_n = a_k + (n - k)d\).
Characteristic property of an arithmetic progression
Any term of the progression, except for the first one, is equal to the arithmetic mean of its neighbors:
This is precisely why the progression is called arithmetic. This equality also works in reverse: if in a sequence each middle term is equal to the average of its neighbors, then the sequence is an arithmetic progression. Such a statement is called a criterion for an arithmetic progression, and it is used in problems to prove that three numbers form a progression.
Example. The numbers 7, \(x\), 23 are three consecutive terms of a progression. Then \(x = \dfrac{7 + 23}{2} = 15\). Check: 7, 15, 23 — the differences are 8 and 8, everything matches.
A convenient way to write the same property is \(2a_n = a_{n-1} + a_{n+1}\), meaning the doubled middle term equals the sum of the outer terms. This form is easier to apply when the terms are given as algebraic expressions.
Sum of the first n terms of an arithmetic progression
The sum \(S_n = a_1 + a_2 + \dots + a_n\) can be calculated using one of two formulas:
The idea behind the first formula is the same one that, according to legend, the young Gauss discovered when summing numbers from 1 to 100. If you add the first term to the last, the second to the second-to-last, and so on, all pairs will yield the same sum. Hence, \(1 + 2 + \dots + 100 = \dfrac{1 + 100}{2}\cdot 100 = 5050\).
The second formula is derived from the first by substituting \(a_n = a_1 + (n-1)d\) — it is useful when the last term is not given and you don't want to calculate it separately.
Example. Let's find the sum of the first twenty terms of the progression 3, 7, 11, 15, 19, …
Using the first formula: \(a_{20} = 79\), so \(S_{20} = \dfrac{3 + 79}{2}\cdot 20 = 41\cdot 20 = 820\).
Using the second formula: \(S_{20} = \dfrac{2\cdot 3 + 19\cdot 4}{2}\cdot 20 = \dfrac{6 + 76}{2}\cdot 20 = 41\cdot 20 = 820\).
The answers match — this is the best way to check calculations.
How to determine if a number belongs to an arithmetic progression
The question of whether a number is a term in a progression is solved by an equation. A number belongs to a progression if and only if its index \(n\) is a natural number: 1, 2, 3, and so on. A fractional or non-positive \(n\) means that such a term does not exist.
Example 1. Does the number 99 belong to the progression 3, 7, 11, 15, 19, …? The formula for the \(n\)-th term here is \(a_n = 4n - 1\), so we solve \(4n - 1 = 99\), which gives \(4n = 100\) and \(n = 25\). The index is a natural number — therefore, 99 is in the progression; it is its twenty-fifth term.
Example 2. What about the number 2026? From \(4n - 1 = 2026\), we get \(4n = 2027\) and \(n = 506.75\). The index is fractional, so 2026 is not a term in this progression.
A useful hint for a quick estimate: for a progression with the formula \(a_n = 4n - 1\), all terms give a remainder of 3 when divided by 4. The number 2026 gives a remainder of 2, so it 'misses' the progression.
Typical OGE problems on arithmetic progression
In the OGE mathematics exam, progressions appear every year, and almost all problems fall into four categories.
Find a term by its index. Given \(a_1\) and \(d\) (or the first few terms) — substitute into \(a_n = a_1 + (n-1)d\).
Find the index by its value. Given a number — equate it to the \(n\)-th term formula and solve the equation for \(n\).
Find the sum. Calculate \(S_n\) using either of the two formulas; if \(a_n\) is missing, find it first.
Find the number of terms. The formula for the \(n\)-th term leads to a convenient consequence: \(n = \dfrac{a_n - a_1}{d} + 1\).
A separate category is word problems, where the progression is hidden within the narrative. For example: on the first day, a worker produced 12 parts, and each subsequent day produced 3 more parts than the previous day; how many parts will he produce in 15 days? Here, \(a_1 = 12\), \(d = 3\), \(n = 15\). Calculate the last term: \(a_{15} = 12 + 14\cdot 3 = 54\). Then \(S_{15} = \dfrac{12 + 54}{2}\cdot 15 = 33\cdot 15 = 495\) parts.
The main advice for such problems: first, write down separately what is given (\(a_1\), \(d\), \(n\), \(a_n\), \(S_n\)) and what needs to be found. Half of the exam errors occur not because of the formulas, but because the index is confused with the term's value.
Arithmetic and geometric progression: what's the difference
Both progressions are rules by which one term is obtained from the next. The difference lies in the operation: in an arithmetic progression, a constant number \(d\) is added to the term, while in a geometric progression, the term is multiplied by a constant number \(q\), called the common ratio.
This leads to different behaviors. An arithmetic progression grows uniformly, 'step by step': 3, 7, 11, 15 — each step adds 4. A geometric progression grows like an avalanche: 3, 6, 12, 24 — each step doubles the number, and after ten steps, the term will be over one and a half thousand.
It's easy to identify the type of sequence: calculate the differences between adjacent terms and their ratios. Equal differences indicate an arithmetic progression, equal ratios indicate a geometric progression. If neither matches, you have a simple sequence: for example, 1, 4, 9, 16 are the squares of natural numbers and not a progression (differences 3, 5, 7 are unequal, as are the ratios).
In the school curriculum, both topics are taught consecutively in the 9th grade, so the formulas can easily be confused. Keep this in mind: in arithmetic progression, the common difference \(d\) is added, while in geometric progression, the common ratio \(q\) is multiplied. Everything else follows from these definitions.
Examples of solving problems on arithmetic progression
Example 1. Find the first term and common difference given two terms
Given. \(a_3 = 14\), \(a_7 = 30\). Find \(a_1\), \(d\), and \(a_{20}\).
Step 1. From the third term to the seventh, there are exactly \(7 - 3 = 4\) steps, with \(d\) added at each step. So, \(a_7 = a_3 + 4d\), which means \(30 = 14 + 4d\), leading to \(4d = 16\) and \(d = 4\).
Step 2. We go back to the first term: \(a_1 = a_3 - 2d = 14 - 8 = 6\).
Step 3. We write the formula for the \(n\)-th term: \(a_n = 6 + (n-1)\cdot 4 = 4n + 2\).
Step 4. We calculate the desired term: \(a_{20} = 4\cdot 20 + 2 = 82\).
Check. Let's list the beginning of the progression: 6, 10, 14, 18, 22, 26, 30. The third term is 14, the seventh is 30 — it matches.
Answer: \(a_1 = 6\), \(d = 4\), \(a_{20} = 82\).
Example 2. Sum of the first 30 terms (both formulas)
Given. Progression \(-5,\ -2,\ 1,\ 4,\ \dots\) Find \(S_{30}\).
Step 1. Determine the parameters: \(a_1 = -5\), \(d = -2 - (-5) = 3\).
Step 2. Find the thirtieth term: \(a_{30} = -5 + 29\cdot 3 = -5 + 87 = 82\).
Step 3. Calculate using the first formula: \(S_{30} = \dfrac{-5 + 82}{2}\cdot 30 = \dfrac{77}{2}\cdot 30 = 77\cdot 15 = 1155\).
Step 4. Check using the second formula. First, the numerator: \(2\cdot(-5) + 29\cdot 3 = -10 + 87 = 77\).
Then, as in step 3: \(S_{30} = \dfrac{77}{2}\cdot 30 = 1155\).
Both methods yielded 1155 — so the arithmetic is correct.
Answer: \(S_{30} = 1155\).
Example 3. Sum of all two-digit numbers divisible by 7
Problem. Find the sum of all two-digit numbers divisible by 7.
Step 1. Such numbers themselves form an arithmetic progression with a common difference of 7. The smallest two-digit multiple of seven is 14, and the largest is 98 (the next one, 105, is already three-digit). We get \(a_1 = 14\), \(d = 7\), \(a_n = 98\).
Step 2. Calculate the number of terms: \(n = \dfrac{98 - 14}{7} + 1 = \dfrac{84}{7} + 1 = 12 + 1 = 13\).
Step 3. Calculate the sum: \(S_{13} = \dfrac{14 + 98}{2}\cdot 13 = \dfrac{112}{2}\cdot 13 = 56\cdot 13 = 728\).
Step 4. Check with the second formula. \(S_{13} = \dfrac{2\cdot 14 + 12\cdot 7}{2}\cdot 13 = \dfrac{28 + 84}{2}\cdot 13 = 56\cdot 13 = 728\).
Answer: 728.
Example 4. Check if a number belongs to a progression
Problem. Is the number 145 a term in the progression \(7,\ 12,\ 17,\ \dots\)? What about the number 142?
Step 1. Parameters of the progression: \(a_1 = 7\), \(d = 5\), so \(a_n = 7 + (n-1)\cdot 5 = 5n + 2\).
Step 2. Check 145: the equation \(5n + 2 = 145\) gives \(5n = 143\) and \(n = 28.6\). The index is fractional, so 145 does not belong to the progression.
Step 3. Check 142: \(5n + 2 = 142\), from which \(5n = 140\) and \(n = 28\). The index is a natural number — the number belongs.
Step 4. Check by substitution. \(a_{28} = 7 + 27\cdot 5 = 7 + 135 = 142\). Correct.
Answer: 145 does not belong to the progression; 142 belongs and is its 28th term.
Example 5. Characteristic property with algebraic terms
Problem. For what value of \(x\) do the numbers \(x - 3\), \(2x + 1\), \(4x - 1\) form three consecutive terms of an arithmetic progression?
Step 1. According to the characteristic property, twice the middle term equals the sum of the outer terms:
Step 2. Expand the parentheses: \(4x + 2 = 5x - 4\).
Step 3. Solve: \(2 + 4 = 5x - 4x\), which means \(x = 6\).
Step 4. Check. Substitute \(x = 6\): the first term is \(6 - 3 = 3\), the second is \(2\cdot 6 + 1 = 13\), the third is \(4\cdot 6 - 1 = 23\). Differences: \(13 - 3 = 10\) and \(23 - 13 = 10\) — they are equal.
Answer: \(x = 6\), the progression is 3, 13, 23 with a common difference of 10.
Example 6. How many negative terms are in the progression
Problem. How many negative terms are there in the progression \(-21,\ -18,\ -15,\ \dots\)?
Step 1. Parameters: \(a_1 = -21\), \(d = 3\). The formula for the \(n\)-th term is: \(a_n = -21 + (n-1)\cdot 3 = 3n - 24\).
Step 2. Write the condition 'term is negative': \(3n - 24 < 0\), which implies \(3n < 24\) and \(n < 8\).
Step 3. Since the indices are natural numbers, \(n = 1, 2, \dots, 7\) are valid — a total of seven terms.
Step 4. Check the boundary. \(a_7 = 3\cdot 7 - 24 = -3\) — negative. \(a_8 = 3\cdot 8 - 24 = 0\) — zero is not a negative number, so the eighth term is not counted. \(a_9 = 3\) — positive.
Answer: seven negative terms.
Common mistakes in arithmetic progression problems
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Using the formula $a_n = a_1 + nd$ instead of $a_n = a_1 + (n-1)d$ for the $n$-th term.
From the first term to the \(n\)-th term, there is one less step than the index. For the progression \(5,\ 9,\ 13,\ \dots\), the tenth term is \(a_{10} = 5 + 9\cdot 4 = 41\), not 45. Quick formula check: substitute \(n = 1\) — you should get exactly \(a_1\).
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Calculating the common difference in reverse: $d = a_1 - a_2$, resulting in the opposite sign.
The common difference is the next term minus the previous term: \(d = a_2 - a_1\). For the progression \(40,\ 34,\ 28,\ \dots\), the correct calculation is \(d = 34 - 40 = -6\). The minus sign here is not an error, but an indicator of a decreasing progression.
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Forgetting to divide by 2 or multiply by $n$ in the sum formula: writing $S_n = (a_1 + a_n)\cdot n$.
Pairs are summed ('first with last'), there are exactly \(\dfrac{n}{2}\) such pairs, so dividing by 2 is essential. For the progression \(2,\ 9,\ 16,\ \dots\), we have \(a_{10} = 2 + 9\cdot 7 = 65\) and \(S_{10} = \dfrac{2 + 65}{2}\cdot 10 = 67\cdot 5 = 335\), not 670.
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Getting $n = 0$ or a negative $n$ when checking if a number belongs to the progression and still answering 'it belongs'.
The index of a term is a natural number; counting starts from one. For example, for the progression \(10,\ 14,\ 18,\ \dots\), the equation \(10 + 4(n-1) = 6\) gives \(n = 0\), which means 6 does not belong to the progression: it would be the term before the first term.
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Confusing the term's value with its index: substituting the term's value where the index is needed, and vice versa.
\(a_n\) is the value, \(n\) is the index. If the problem asks 'which term number equals 142', the answer is 28, not 142. A useful habit: before solving, write down 'Given: \(a_1 = \dots\), \(d = \dots\), Find: \(n\)'.
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Assuming that a progression must be increasing and consist of integers.
The common difference can be negative (\(20,\ 17,\ 14,\ \dots\)), fractional (1; 1.5; 2; 2.5), or even zero (8, 8, 8, …). All three cases are valid arithmetic progressions, and all formulas work for them without changes.
Questions and Answers
What is an arithmetic progression in simple terms?
It's a sequence of numbers where each subsequent number is obtained by adding the same number \(d\) — the common difference — to the previous one. For example, 3, 7, 11, 15, 19, … is a progression with the first term 3 and a common difference of 4.
What is the formula for the n-th term of an arithmetic progression?
\(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the index. If not the first term, but some other term is known, the general form \(a_n = a_k + (n-k)d\) applies.
How to find the common difference of an arithmetic progression?
Subtract the previous term from any term: \(d = a_n - a_{n-1}\). If two terms with different indices are given, use \(d = \dfrac{a_k - a_m}{k-m}\). For example, with \(a_5 = 23\) and \(a_9 = 43\), we get \(d = \dfrac{43-23}{9-5} = 5\), and the first term is \(23 - 4\cdot 5 = 3\).
How to find the first term of an arithmetic progression?
From the formula for the \(n\)-th term: \(a_1 = a_n - (n-1)d\). For example, if \(a_{12} = 60\) and \(d = 4\), then \(a_1 = 60 - 11\cdot 4 = 60 - 44 = 16\). Check: \(16 + 11\cdot 4 = 60\).
How to find the sum of the first n terms of an arithmetic progression?
Use the formula \(S_n = \dfrac{a_1 + a_n}{2}\cdot n\) if the last term is known, or \(S_n = \dfrac{2a_1 + (n-1)d}{2}\cdot n\) if only the first term and common difference are known. Both give the same result, so the second can be used to check the first.
How to find the number of terms in an arithmetic progression?
From the formula for the \(n\)-th term: \(n = \dfrac{a_n - a_1}{d} + 1\). For example, in the progression \(8,\ 13,\ 18,\ \dots,\ 98\), the common difference is 5, so \(n = \dfrac{98-8}{5} + 1 = 18 + 1 = 19\) terms.
How does an arithmetic progression differ from a geometric progression?
In an arithmetic progression, the common difference \(d\) is added to each term, while in a geometric progression, each term is multiplied by the common ratio \(q\). The formulas for the \(n\)-th term are: \(a_n = a_1 + (n-1)d\) versus \(b_n = b_1 q^{\,n-1}\). To determine the type, calculate the differences between adjacent terms and their ratios: equal differences mean arithmetic progression, equal ratios mean geometric progression.
Can the common difference be negative or zero?
Yes. If \(d < 0\), the progression is decreasing (e.g., 20, 17, 14, 11). If \(d = 0\), all terms are the same (6, 6, 6, …) — this is a constant progression. The common difference can also be fractional: 1; 1.5; 2; 2.5 is also an arithmetic progression.
What grade is an arithmetic progression studied in?
It is studied in the 9th grade in algebra classes, along with geometric progressions and numerical sequences. Problems on progressions are included in the OGE: typically, you need to find a term by its index, an index by its value, the number of terms, or the sum of the first \(n\) terms.
