CAT 2006 — QA Question 12
Arithmetic ProgressionEasy
Consider the set S = {1, 2, 3, …, 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?
Answer & solution
- A
3
- B
4
- C
6
7
- E
8
Solution
Let there be n terms (n ≥ 3) in the arithmetic progression having 1 as the first term and 1000 as the last. Let d be the common difference. Then,
1000 = 1 + (n – 1) × d
∴ 999 = (n – 1) × d ... (i)
∴ Factors of 999 are 1, 3, 9, 27, 37, 111, 333 and 999
Substituting in equation (i)
If d = 1, n = 1000
If d = 3, n = 334
If d = 9, n = 112
If d = 27, n = 38
If d = 37, n = 28
If d = 111, n = 10
If d = 333, n = 4
If d = 999, n = 2, which is not possible as n > 2
∴ 7 arithmetic progressions can be formed.
Hence, option (d).