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?
- (i) 2, 5, 8, 11, …
- (ii) −5, −1, 3, 7, …
Step-by-step solution
Idea: An AP has a constant difference d, and its nth term is a + (n − 1)d. Plotting (position, term) for an AP always gives points on a straight line, because each step right by 1 goes up by the same amount d.
(i) 2, 5, 8, 11, …
- Differences: 5 − 2 = 3, 8 − 5 = 3, 11 − 8 = 3. Constant, so it is an AP with a = 2, d = 3.½ mark
- tn = 2 + (n − 1) × 3 = 3n − 1.½ mark
- Points (1, 2), (2, 5), (3, 8), (4, 11): each step right by 1 goes up by 3, so they lie on a straight line (blue line in the graph).½ mark
(ii) −5, −1, 3, 7, …
- Differences: −1 − (−5) = 4, 3 − (−1) = 4, 7 − 3 = 4. Constant, so it is an AP with a = −5, d = 4.½ mark
- tn = −5 + (n − 1) × 4 = 4n − 9.½ mark
- Points (1, −5), (2, −1), (3, 3), (4, 7): each step right by 1 goes up by 4, so they also lie on a straight line, a steeper one (red line).½ mark
Check: (i) 3 × 4 − 1 = 11 ✓ (ii) 4 × 4 − 9 = 7 ✓.
Answer to write in the exam
(i)
5 − 2 = 8 − 5 = 11 − 8 = 3 (constant) ∴ AP; a = 2, d = 3
tn = a + (n − 1)d = 2 + 3(n − 1)
∴ tn = 3n − 1
Points (1, 2), (2, 5), (3, 8), (4, 11) lie on a straight line.
(ii)
−1 − (−5) = 3 − (−1) = 7 − 3 = 4 (constant) ∴ AP; a = −5, d = 4
tn = −5 + 4(n − 1)
∴ tn = 4n − 9
Points (1, −5), (2, −1), (3, 3), (4, 7) lie on a straight line.
Common mistakes that cost marks
- In (ii), computing −1 − (−5) as −6. Subtracting a negative adds: −1 + 5 = 4.
- Writing the nth term as a + nd (e.g. 2 + 3n). It must give the first term when n = 1, so use a + (n − 1)d.
- Joining the points with a curve or plotting (term, position) the wrong way round. Plot (n, tn) with n on the x-axis.
How this can come in the exam
The nth term of the AP −2, 4, 10, 16, … is
- 6n − 2
- 6n − 8
- 4n − 6
- −2n + 6
Show answer
(B) 6n − 8
a = −2, d = 6: −2 + 6(n − 1) = 6n − 8.
Assertion (A): The points (1, 7), (2, 10), (3, 13), (4, 16) lie on a straight line.
Reason (R): 7, 10, 13, 16 is an AP.
- 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.
The terms go up by the same 3 for each step of 1 in n, so the points of an AP lie on a straight line. R explains A.
Try one yourself
Show that 6, 4, 2, 0, … is an AP, find its nth term, and describe its graph.
Show answer
Differences are all −2, so it is an AP. tn = 6 − 2(n − 1) = 8 − 2n. The points (1, 6), (2, 4), (3, 2), (4, 0) lie on a straight line going down.
More questions like this
- 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?
- Can you find the sum of the first ten natural numbers without actually adding all of them?
All Sequences and progressions questions · All maths questions