The sequence (m+1, 3m−2, 6m−8) is stated not to be an arithmetic progression. For which value of m will this statement become false?

Correct answer: 3

The statement becomes false precisely when the three displayed terms do form an arithmetic progression. For three terms, apply the condition that twice the middle term equals the sum of the first and third terms: 2(3m − 2) = (m + 1) + (6m − 8). This gives 6m − 4 = 7m − 7, and hence m = 3. Substitution confirms the result: the terms are 4, 7, and 10, whose consecutive differences are both 3. Thus the original claim that the sequence is not an AP is false, and option D is correct. The values 1, 2, and 4 do not produce equal consecutive differences. The phrase “become false” is important: we are seeking the value that makes the AP condition true.