Using the formula tn = a + (n − 1) × d, find the nth term of the following arithmetic progressions.
- (i) 12, 52, 92, 132, …
- (ii) 1.5, 3.5, 5.5, 7.5, …
Answer: (i) tn = 2n − 32 = 4n − 32 (ii) tn = 2n − 0.5
Step-by-step solution
Idea: Read off the first term a and the common difference d (any term minus the one before), then put them into a + (n − 1)d and simplify.
(i) 12, 52, 92, 132, …
- a = 12, d = 52 − 12 = 42 = 2.½ mark
- tn = 12 + (n − 1) × 2 = 2n − 2 + 12 = 2n − 32 = 4n − 32.½ mark
2n − 32, that is, 4n − 32
(ii) 1.5, 3.5, 5.5, 7.5, …
- a = 1.5, d = 3.5 − 1.5 = 2.½ mark
- tn = 1.5 + (n − 1) × 2 = 1.5 + 2n − 2 = 2n − 0.5.½ mark
2n − 0.5
(i) tₙ = 2n − 3/2 = (4n − 3)/2 (ii) tₙ = 2n − 0.5
Check: (i) n = 4: (16 − 3)/2 = 13/2 ✓ (ii) n = 4: 8 − 0.5 = 7.5 ✓.
Answer to write in the exam
(i)
a = 12, d = 52 − 12 = 2
tn = 12 + (n − 1) × 2
∴ tn = 2n − 32 = 4n − 32
(ii)
a = 1.5, d = 3.5 − 1.5 = 2
tn = 1.5 + (n − 1) × 2
∴ tn = 2n − 0.5
Common mistakes that cost marks
- In (i), taking d = 42 and leaving it there, or taking d = 4 by subtracting only the numerators and forgetting the denominator.
- Expanding (n − 1) × 2 as 2n − 1. It is 2n − 2.
- Writing 1.5 + 2n − 2 = 2n + 0.5. Since 1.5 − 2 = −0.5, the answer is 2n − 0.5.
How this can come in the exam
MCQ (1 mark)
The nth term of the AP 13, 1, 53, 73, … is
- 2n − 13
- n + 23
- 2n3
- 3n − 23
Show answer
(A) 2n − 13
a = 13, d = 23: 13 + 23(n − 1) = 2n − 13.
Try one yourself
Find the nth term of the AP 0.25, 0.75, 1.25, 1.75, …
Show answer
a = 0.25, d = 0.5: tn = 0.25 + 0.5(n − 1) = 0.5n − 0.25.
More questions like this
- 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?
- Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?
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