Three distinct positive integers a, b and c are such that b – a = c – b and a + b + c = 12. What is the maximum number of such possible sets (a, b, c) ?

Correct — C, (c) 3. There are exactly three such sets: (1, 4, 7), (2, 4, 6) and (3, 4, 5).

Read the first condition. b − a = c − b says the gap from a to b equals the gap from b to c — the three numbers are in ARITHMETIC PROGRESSION, with b in the middle. Rearranged, it gives 2b = a + c.

Use the second condition. a + b + c = 12. Substituting a + c = 2b gives 2b + b = 12, so 3b = 12 and b = 4. The middle number is fixed before anything else is decided. That is the key step, and it is worth remembering as a general fact: for three numbers in arithmetic progression, the middle one is always the average of the three.

Now count the possibilities. Write a = 4 − d and c = 4 + d, where d is the common difference. Two conditions restrict d: — the integers must be DISTINCT, so d cannot be 0; — all three must be POSITIVE, so a = 4 − d must be at least 1, giving d at most 3. So d = 1, 2 or 3, and the sets are (3, 4, 5), (2, 4, 6) and (1, 4, 7). Each sums to 12 and each is in arithmetic progression. Three sets.

One point about the counting. Writing the same three numbers in the reverse order — (5, 4, 3), say — also satisfies b − a = c − b, since both differences become negative. But that is the same SET of numbers, and the question asks for the number of sets; counting orderings as well would give six, which is not among the options and is not what the question asks. The distinctness requirement is what keeps (4, 4, 4) out, and the positivity requirement is what keeps (0, 4, 8) and (−1, 4, 9) out.

Why the others are wrong

• (a)1 — One is the count if only CONSECUTIVE integers are considered — the set (3, 4, 5). But arithmetic progression does not mean consecutive; it means a constant gap, and gaps of 2 and 3 work equally well here, giving (2, 4, 6) and (1, 4, 7). Reading 'b − a = c − b' as 'the numbers follow one another' is the error, and it is common.
• (b)2 — Two is the count if one of the three admissible common differences is overlooked, most often d = 3, which produces (1, 4, 7). That set is easy to miss because its first term is the smallest positive integer and it sits right at the boundary of the constraint. Since a = 4 − d must be at least 1, d = 3 is allowed, and the boundary case is always worth writing out explicitly.
• (d)4 — Four is the count if d = 0 is admitted, giving (4, 4, 4) alongside the three genuine sets. That triple does satisfy both equations — its differences are equal and its sum is 12 — but the stem requires the integers to be DISTINCT, which rules it out. In problems of this kind the word 'distinct' is always doing work and is easy to skim past.

Three numbers in arithmetic progression are most conveniently written as m − d, m and m + d, because the outer terms then cancel when they are added: the sum is 3m, so the middle term is always the mean. The same device extends — four terms in arithmetic progression are written m − 3d, m − d, m + d, m + 3d so that their sum is 4m. Counting problems built on such progressions are then a matter of finding which values of the common difference satisfy the stated restrictions, and the restrictions are usually of two kinds: distinctness, which excludes d = 0, and positivity or a range, which caps how large d can be. Listing the qualifying cases explicitly, rather than counting them in your head, is what prevents an off-by-one answer.

Key facts

• b − a = c − b means a, b and c are in arithmetic progression.
• For three terms in arithmetic progression, the middle term is the average: 2b = a + c.
• With a + b + c = 12, it follows that 3b = 12 and b = 4.
• Writing a = 4 − d and c = 4 + d, distinctness requires d ≠ 0 and positivity requires d ≤ 3.
• The admissible sets are (3, 4, 5), (2, 4, 6) and (1, 4, 7) — three in all.
• (4, 4, 4) satisfies both equations but fails the distinctness requirement.
• Reversing the order of a set does not create a new set.
• Three terms in arithmetic progression are best written as m − d, m, m + d; four as m − 3d, m − d, m + d, m + 3d.

Common traps

• Reading arithmetic progression as consecutive integers.
• Missing the boundary case where the smallest term equals 1.
• Admitting a common difference of zero despite the word 'distinct'.
• Counting orderings of the same three numbers as separate sets.

• practice — not a real PYQ

  Three distinct positive integers in arithmetic progression have a sum of 18. How many such sets are possible ?

  • (a)3
  • (b)4
  • (c)5
  • (d)6

  Answer(c) 5

• practice — not a real PYQ

  If three numbers are in arithmetic progression and their sum is 45, what is the middle number ?

  • (a)9
  • (b)12
  • (c)15
  • (d)It cannot be determined

  Answer(c) 15