Arithmetic Progressions
CBSE Class 10 · Mathematics · Notes, formulas and practice questions
An arithmetic progression is a sequence where each term differs from the previous one by the same fixed number. This chapter explains how to find any term and the sum of the first n terms, and how to use these ideas in everyday problems.
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An arithmetic progression is a list of numbers in which the difference between one term and the next is constant throughout. Working with an AP begins by identifying its first term, written as a, and this fixed common difference, written as d. These two values tell you whether the list increases, decreases, or stays the same, and every later formula in the chapter is built on them. A sequence is an AP only if every pair of consecutive terms gives the same d, so checking this difference should be the first step in any problem.
The direct way to locate any term of an AP is the general term aₙ = a + (n − 1)d. Rather than memorising it mechanically, notice that this formula counts the first term plus d added once for the second term, twice for the third term, and so on. This one rule answers many questions: finding a specified term, deciding which term has a given value, and counting how many terms an AP contains. When two different terms are given, substituting each into the general term produces two equations, from which a and d can be found.
For many problems the total of the first n terms is needed, not just a single term. By writing the sum from the first to the last term and then backward, each pair adds to the same value, which leads to Sₙ = (n/2)[2a + (n − 1)d]. If the last term in the group is known, the shorter form Sₙ = (n/2)(a + l) is also useful. Setting a = 1 and d = 1 in the same rule gives the sum of the first n natural numbers, n(n + 1)/2.
Real-life situations that change by equal amounts at each step, such as saving money, paying equal instalments, arranging seats in rows, or rising or falling by a fixed amount, can be modelled as APs. To solve such a problem, read the initial value as a, the constant step as d, a particular moment as n, and a total amount as Sₙ. A common mistake is to use the nth term formula when the question asks for a total sum, so identify whether a single term or a sum is required before substituting. Another useful relation is aₙ = Sₙ − Sₙ₋₁, which lets you find the general term when the sum formula is given.
Key terms
- Sequence
- An ordered list of numbers, where each number is called a term. The first number is the first term and the number at position n is generally written as aₙ.
- Arithmetic progression (AP)
- A sequence in which each term after the first is obtained by adding the same fixed number to the previous term. For example, 3, 7, 11, 15 is an AP because each term is 4 more than the previous one.
- First term
- The starting value of an arithmetic progression, usually denoted by a. In the AP 5, 8, 11, 14, the first term is a = 5.
- Common difference
- The constant difference between consecutive terms of an AP, denoted by d. It is found by subtracting a term from the next term, so d = a₂ − a₁ = a₃ − a₂, and it may be positive, negative, or zero.
- nth term (general term)
- The term in position n of an AP, given by aₙ = a + (n − 1)d. This formula lets you find any term without writing out the entire progression.
- Sum of the first n terms
- The total obtained by adding the first n terms of an AP, written Sₙ. It is calculated using Sₙ = (n/2)[2a + (n − 1)d] or, when the last term l is known, Sₙ = (n/2)(a + l).
- Increasing, decreasing, and constant AP
- If d > 0, the terms become larger; if d < 0, the terms become smaller; if d = 0, all terms stay equal to a. The value of d is the same for every pair of consecutive terms.
- Real-life model as an AP
- A real situation can be modelled by an AP when a quantity starts at a and changes by the same amount after every equal step. The general term predicts the quantity at a chosen step, and Sₙ gives its total over n steps.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| General term (nth term) | aₙ = a + (n − 1)d | a is the first term, d is the common difference, and n is the position of the term (n = 1, 2, 3, …). |
| Common difference | d = a₂ − a₁ | a₁ and a₂ are any two consecutive terms; in an AP this difference is the same for every consecutive pair. |
| Sum of the first n terms | Sₙ = (n/2)[2a + (n − 1)d] | Sₙ is the sum of the first n terms, a is the first term, d is the common difference, and n is a positive integer. |
| Sum using the first and last term | Sₙ = (n/2)(a + l) | a is the first term, l is the last of the n terms being added, and n is the number of terms in the sum. |
| General term from the sum | aₙ = Sₙ − Sₙ₋₁ | Sₙ is the sum of the first n terms and Sₙ₋₁ is the sum of the first n − 1 terms; this gives the nth term directly. |
| Sum of the first n natural numbers | 1 + 2 + 3 + … + n = n(n + 1)/2 | This is the special AP with a = 1 and d = 1; n is any positive integer. |
Practice questions with answers
1. Find the 10th term of the AP: 2, 7, 12, 17, …
The first term is a = 2 and the common difference is d = 7 − 2 = 5. Using aₙ = a + (n − 1)d gives a₁₀ = 2 + 9 × 5 = 47. So the 10th term is 47.
2. Which term of the AP: 21, 18, 15, … is −81?
Here a = 21 and d = 18 − 21 = −3. Let the required term be the nth term, so 21 + (n − 1)(−3) = −81. Dividing through gives n − 1 = 34, so n = 35. Therefore −81 is the 35th term.
3. Write the first four terms of an AP whose first term is −2 and common difference is 5.
Start with a = −2 and add d = 5 repeatedly. The terms are −2, 3, 8, and 13, since −2 + 5 = 3, 3 + 5 = 8, and 8 + 5 = 13.
4. Determine the AP whose 3rd term is 5 and 7th term is 9.
Let a be the first term and d the common difference. Then a + 2d = 5 and a + 6d = 9. Subtracting gives 4d = 4, so d = 1, and substituting back gives a = 3. Therefore the AP is 3, 4, 5, 6, …
5. How many terms of the AP: 24, 21, 18, … must be taken so that their sum is 78?
Let the number of terms be n. Here a = 24 and d = −3, so Sₙ = (n/2)[48 + (n − 1)(−3)] = n(51 − 3n)/2. Setting Sₙ = 78 gives n² − 17n + 52 = 0, so n = 4 or n = 13. Both answers are valid because the first four terms add to 78 and the next nine terms together add to zero.
6. A person saves ₹250 in the first week and increases the weekly saving by ₹50 each week. What is the total saving in the first 12 weeks?
The weekly savings form an AP with a = 250 and d = 50. S₁₂ = (12/2)[2 × 250 + 11 × 50] = 6 × [500 + 550] = 6 × 1050 = 6300. Hence the total saving in the first 12 weeks is ₹6300.
7. Find the sum of the first 20 positive odd integers.
The positive odd integers are 1, 3, 5, …, which form an AP with a = 1 and d = 2. Using Sₙ = (n/2)[2a + (n − 1)d], S₂₀ = 10 × [2 + 19 × 2] = 10 × 40 = 400. So the required sum is 400.
8. Find the sum of the first 100 positive multiples of 5.
The multiples are 5, 10, 15, …, 500. This is an AP with a = 5 and l = 500. S₁₀₀ = (100/2)(5 + 500) = 50 × 505 = 25250.
9. The sum of the first n terms of an AP is 2n² + 3n. Find its nth term.
Use aₙ = Sₙ − Sₙ₋₁. Since Sₙ = 2n² + 3n, substitute n − 1 to get Sₙ₋₁ = 2(n − 1)² + 3(n − 1) = 2n² − n − 1. Therefore aₙ = (2n² + 3n) − (2n² − n − 1) = 4n + 1. The first term is 5 and the common difference is 4, which checks with the given sum.
10. State whether the following sequences are APs, giving reasons: (i) −3, −7, −11, −15; (ii) 1, 4, 9, 16.
For (i), the consecutive differences are −4, −4, and −4, which are equal, so it is an AP with a = −3 and d = −4. For (ii), the differences are 3, 5, and 7, which are not the same, so it is not an AP.
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