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Arithmetic progressions: definitions and examples

A sequence is a list of numbers a1,a2,a3,a_1, a_2, a_3, \ldots written in some specific order. The number ana_n is called the nnth term of the sequence. Sequences can follow any rule , squares, primes, Fibonacci-like recurrences , but this chapter focuses on the simplest interesting rule: each term is obtained by adding a fixed number to the previous term.

Definitions

A sequence a1,a2,a3,a_1, a_2, a_3, \ldots is an arithmetic progression (AP) if there exists a number dd such that an+1an=dfor every n1.a_{n+1} - a_n = d \quad \text{for every } n \ge 1.

This dd is called the common difference. The first term is denoted aa (or a1a_1). An AP is completely specified by the pair (a,d)(a, d): a,a+d,a+2d,a+3d,a, \quad a + d, \quad a + 2 d, \quad a + 3 d, \quad \ldots

A finite AP has a fixed number of terms (nn terms). Its last term is denoted =a+(n1)d\ell = a + (n - 1) d.

A sequence is an AP iff every consecutive difference is the same. The test is: compute a2a1a_2 - a_1, a3a2a_3 - a_2, a4a3a_4 - a_3, etc. If they all match, it is an AP with that common value as dd. If any differ, it is not.

Recognising APs

Examples and non-examples:

  • 3,7,11,15,19,3, 7, 11, 15, 19, \ldots , differences are all 44. AP, a=3,d=4a = 3, d = 4.
  • 5,2,1,4,7,5, 2, -1, -4, -7, \ldots , differences are 3-3 throughout. AP, a=5,d=3a = 5, d = -3.
  • 1,4,9,16,25,1, 4, 9, 16, 25, \ldots , differences are 3,5,7,93, 5, 7, 9 (changing). Not an AP.
  • 2,4,8,16,32,2, 4, 8, 16, 32, \ldots , differences are 2,4,8,162, 4, 8, 16 (changing). Not an AP (this is a geometric progression).
  • 10,10,10,10-10, -10, -10, -10 , common difference 00. AP with d=0d = 0 (a constant sequence).
  • 2,8,18,32,\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots , simplifying: 2,22,32,42,\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \ldots , differences 2\sqrt{2}. AP.

Some real-life APs:

  • The cost of a taxi: ₹1515 for the first km, then ₹88 for each subsequent km , fares: 15,23,31,15, 23, 31, \ldots. AP with a=15,d=8a = 15, d = 8.
  • Each row of an auditorium has 22 more seats than the row in front. Seats per row form an AP.
  • A salary scheme with a fixed annual increment.

Theorem / Concept: characterising APs

Test for an AP. A list a1,a2,a_1, a_2, \ldots is an AP if and only if there is a constant dd with an+1an=da_{n+1} - a_n = d for all nn. Equivalently, 2an+1=an+an+22 a_{n+1} = a_n + a_{n+2} for all nn (the middle-term test).

The middle-term test is often easier in problems where you are given three numbers and asked to find a parameter that makes them an AP.

If the sequence is given by a formula (like an=3n+5a_n = 3 n + 5), you can read off dd by computing an+1ana_{n+1} - a_n. In general, ana_n is a linear function of nn , an=a+(n1)d=dn+(ad)a_n = a + (n - 1) d = d n + (a - d).

So an AP is exactly a sequence whose general term is linear in nn. This is one of the cleanest characterisations and worth remembering.

Worked examples

Example 1. Find the first term and common difference of 7,13,19,25,7, 13, 19, 25, \ldots.

a=7a = 7. Differences: 137=613 - 7 = 6, 1913=619 - 13 = 6, 2519=625 - 19 = 6. So d=6d = 6.

Example 2. Is 10,6,2,2-10, -6, -2, 2 an AP? If so, find the next two terms.

Differences: 6(10)=4-6 - (-10) = 4, 2(6)=4-2 - (-6) = 4, 2(2)=42 - (-2) = 4. Yes, AP with a=10,d=4a = -10, d = 4.

Next two terms: 2+4=62 + 4 = 6 and 6+4=106 + 4 = 10.

Example 3. For what value of kk are kk, 2k12 k - 1, 2k+12 k + 1 in AP?

Middle term test: 2(2k1)=k+(2k+1)4k2=3k+1k=32(2 k - 1) = k + (2 k + 1) \Rightarrow 4 k - 2 = 3 k + 1 \Rightarrow k = 3.

Example 4. The nnth term of a sequence is an=3n+1a_n = 3 n + 1. Show it is an AP and find aa and dd.

an=3n+1a_n = 3 n + 1 is linear in nn. Differences: an+1an=3(n+1)+1(3n+1)=3a_{n+1} - a_n = 3(n+1) + 1 - (3 n + 1) = 3. So AP with d=3d = 3, and a=a1=4a = a_1 = 4.

Example 5. The first three terms of an AP are 2x+1,3x+2,2x+52 x + 1, 3 x + 2, 2 x + 5. Find xx.

By the middle-term test: 2(3x+2)=(2x+1)+(2x+5)6x+4=4x+6x=12 (3 x + 2) = (2 x + 1) + (2 x + 5) \Rightarrow 6 x + 4 = 4 x + 6 \Rightarrow x = 1.

Check: terms become 3,5,73, 5, 7. Common difference 22. ✓

Try it yourself

  1. Which of these are APs? For each AP, find aa and dd: (i) 11,9,7,5,11, 9, 7, 5, \ldots, (ii) 2,6,12,20,2, 6, 12, 20, \ldots, (iii) 0.5,1,1.5,2,0.5, 1, 1.5, 2, \ldots, (iv) 3,12,27,48,\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots, (v) 1,1,1,1,1, 1, 1, 1, \ldots.
  2. Find xx if x2,2x,3x+1x - 2, 2 x, 3 x + 1 are in AP.
  3. Find the first term and common difference of an AP whose 55th term is 1919 and 99th term is 3535.
  4. Write the first four terms of the AP with a=2,d=0a = -2, d = 0.
  5. Does the sequence an=42na_n = 4 - 2 n form an AP? If yes, find dd.
  6. Find kk so that k+9,2k1,2k+7k + 9, 2 k - 1, 2 k + 7 form an AP.
  7. Are 1,4,9,161, 4, 9, 16 an AP? Justify.
  8. Find the 2020th term of 5,12,19,26,5, 12, 19, 26, \ldots.
  9. The nnth term of an AP is 53n5 - 3 n. Find aa and dd.
  10. Three consecutive terms of an AP are a5,a,a+5a - 5, a, a + 5. Verify by the test.

Pitfalls / Insight

  • Always check the difference for the first pair AND a later pair. Sometimes only the first few terms look like an AP; later they don't.
  • dd can be negative or zero. A constant sequence is a valid AP.
  • Don't confuse AP with GP. Multiplying by a fixed factor each time is geometric, not arithmetic.

Insight. Identifying an AP is almost always the easy step. The real work is finding aa and dd from the information you have. Two pieces of information are typically enough , use them to set up two equations in a,da, d.

Practice quiz

Quick check on this topic.

Quiz
Quick check : AP basics
6 questions · pick the best answer
Q1

11,9,7,511, 9, 7, 5 is an AP with d=d = :

Q2

2,8,18,32\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} is:

Q3

1,4,9,16,1, 4, 9, 16, \ldots is:

Q4

If an=2n+3a_n = 2n + 3, then d=d = :

Q5

Three numbers k,2k1,2k+1k, 2k - 1, 2k + 1 in AP. k=k = :

Q6

The sequence 10,10,10,-10, -10, -10, \ldots is: