Consider all the sequences we have discussed so far. Which ones are arithmetic progressions and which ones are not? Can you justify your claim?
Step-by-step solution
Idea: A sequence is an AP when the difference between every pair of consecutive terms is the same. To show a sequence is not an AP, one pair of unequal differences is enough.
- Test: work out (term − previous term) for several pairs. Same every time ⇒ AP; changes ⇒ not an AP.½ mark
- APs: natural numbers 1, 2, 3, … (difference 1); odd numbers 1, 3, 5, … (difference 2; this is also 2n − 1); 1, 4, 7, 10, 13, … (difference 3).1 mark
- Also APs: −7, −3, 1, 5, 9, … (difference 4); 5n − 2: 3, 8, 13, … (difference 5); 3n − 7: −4, −1, 2, … (difference 3); the squares pattern 1, 5, 9, 13, … (difference 4). From the exercises: 3n − 4: −1, 2, 5, … (difference 3); 2 − 5n: −3, −8, −13, … (difference −5); 5n − 3: 2, 7, 12, … (difference 5); −5, −2, 1, 4, … (difference 3). Any rule of the form (number) × n + (number) gives an AP.1 mark
- Not APs: triangular numbers 1, 3, 6, 10 (differences 2, 3, 4); square numbers 1, 4, 9, 16 (differences 3, 5, 7); 6, 12, 24, 48, 96 (differences 6, 12, 24); 1, 12, 13, 14 (differences −12, −16, …).1 mark
- Also not APs: the primes 2, 3, 5, 7 (differences 1, 2, 2); 1, 5, 13, 29 (differences 4, 8, 16); 3, 6, 30, 870; the Virahānka–Fibonacci sequence 1, 2, 3, 5, 8 (differences 1, 1, 2, 3); n2 − 2n + 3: 2, 3, 6, 11 (differences 1, 3, 5); the running sums 1, 5, 12, 22 (differences 4, 7, 10); and 1, 2, 4, 7, 13 (differences 1, 2, 3, 6).½ mark
Answer to write in the exam
AP ⇔ (term − previous term) is the same throughout.
APs: 1, 2, 3, … (d = 1); 1, 3, 5, … (d = 2); 1, 4, 7, 10, … (d = 3); −7, −3, 1, 5, … (d = 4); 5n − 2 (d = 5); 3n − 7 (d = 3); 1, 5, 9, 13, … (d = 4); 3n − 4 (d = 3); 2 − 5n (d = −5); 5n − 3 (d = 5); −5, −2, 1, 4, … (d = 3)
Not APs: triangular numbers (differences 2, 3, 4, …); square numbers (3, 5, 7, …); 6, 12, 24, 48, 96 (6, 12, 24, …); 1, 12, 13, … (−12, −16, …)
Not APs: primes (1, 2, 2, …); 1, 5, 13, 29 (4, 8, 16); 3, 6, 30, 870; 1, 2, 3, 5, 8, … (1, 1, 2, …); n2 − 2n + 3: 2, 3, 6, 11 (1, 3, 5); 1, 5, 12, 22 (4, 7, 10); 1, 2, 4, 7, 13 (1, 2, 3, 6)
∴ Only the sequences with a constant difference are APs.
Common mistakes that cost marks
- Calling a sequence an AP because it increases. Square numbers increase but their differences (3, 5, 7, …) are not constant.
- Checking only the first difference. 1, 2, 4, 7, … starts with difference 1 but the next is 2.
- Thinking a decreasing sequence cannot be an AP. 11, 7, 3, −1, … is an AP with difference −4.
How this can come in the exam
Which of these is an arithmetic progression?
- 1, 4, 9, 16, …
- 2, 4, 8, 16, …
- 10, 7, 4, 1, …
- 1, 2, 3, 5, 8, …
Show answer
(C) 10, 7, 4, 1, …
10, 7, 4, 1 has a constant difference of −3.
Assertion (A): 1, 3, 6, 10, 15, … is not an AP.
Reason (R): The differences between consecutive terms are 2, 3, 4, 5, which are not all equal.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(A) Both A and R are true, and R is the correct explanation of A.
An AP needs equal differences; R shows they are not equal, which is exactly why A is true.
Try one yourself
Which of these are APs? (a) 4, 9, 14, 19, … (b) 3, 9, 27, 81, … (c) 0, −2, −4, −6, …
Show answer
(a) AP, difference 5. (b) Not an AP: differences 6, 18, 54. (c) AP, difference −2.
More questions like this
- Verify that the following sequences are arithmetic progressions and write their nth terms. What do you observe when you plot the ordered pairs emerging from them?
- Using the formula tn = a + (n − 1) × d, find the nth term of the following arithmetic progressions.
- Consider once again the AP: 1, 5, 9, 13, 17, … Since every term is 4 more than the previous term, another way of writing this sequence is t1 = 1, tn = tn−1 + 4 for n ≥ 2. Verify for yourself that this recursive rule leads to the terms of the sequence.
- Find recursive rules for the APs in the previous exercises.
- A person books a taxi to travel in the city. The taxi company charges a fixed booking fee of ₹200 plus ₹40 per kilometre travelled. Let us write the sequence representing the total fare after travelling 1 km, 2 km, 3 km, and so on. If the person travels 10 km, what will be the total fare?
All Sequences and progressions questions · All maths questions