Discussion

Explanation:

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).

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