In a race, the first four winners are to be awarded points. Each winner's points must be 5 more than that of the next position winner. Total sum of the points to be awarded is 50. What will be the points for the third position winner ?

Correct — C, (c) 10.

Four awards, each 5 above the one below it, is an ARITHMETIC PROGRESSION of four terms with a common difference of 5. Name the smallest and the rest follow.

fourth place a third place a + 5 second place a + 10 first place a + 15

sum = 4a + 30 = 50 4a = 20, so a = 5

The third position gets a + 5 = 10, which is option (c). The full award is 20, 15, 10 and 5, and those four do sum to 50.

THE FASTER ROUTE, and the one worth owning for every progression question, uses the mean. In an arithmetic progression the terms are evenly spaced, so their average is the average of the whole run — and the average here is 50 ÷ 4 = 12·5. With an EVEN number of terms the average sits exactly halfway between the two middle ones, so

second place + third place = 2 × 12·5 = 25

and since the second is 5 above the third, 2 × (third) + 5 = 25, giving the third place 10 immediately. No unknown has to be introduced at all.

WITH AN ODD NUMBER OF TERMS the shortcut is even shorter: the average IS the middle term. Five prizes in progression totalling 5000 make the third prize 1000 without any working.

WHERE THE STEM CAN BE MISREAD. 'Each winner's points must be 5 more than that of the next position winner' fixes the DIRECTION: the higher position has more, and the values descend as the positions descend. Reading it the other way round produces the same four numbers attached to the opposite positions, and the paper offers those numbers.

A CHECK THAT COSTS NOTHING. Once you have a candidate value, rebuild the whole sequence around it and add. If the third place were 30, the four awards would be 40, 35, 30 and 25, totalling 130 rather than 50. If it were 20, they would be 30, 25, 20 and 15, totalling 90. Only 10 reconstructs to a total of 50.

The options are printed as bare numerals with no unit attached.

Why the others are wrong

• (a)30 — The only offered value that is not a score in the sequence at all, and it is refuted by rebuilding the run around it: a third place of 30 makes the four awards 40, 35, 30 and 25, which total 130 and not the 50 the question states. Where it comes from is the equation itself. Setting the smallest award as a gives 4a + 30 = 50, and the 30 in that line is the pooled effect of the three increments, 5 + 10 + 15. A candidate who assembles the equation correctly and then reads the wrong number off it lands here. The lesson is that a constant appearing inside a working is not a candidate answer, and rebuilding the sequence is a two-second test that catches it.
• (b)20 — A real score in the sequence — the FIRST position's award — offered to a candidate who solves the problem correctly and then reads off the wrong end of the run. The four awards are 20, 15, 10 and 5 for the first, second, third and fourth positions, so a candidate who finds the largest and stops, or who counts the positions from the bottom instead of from the top, chooses 20. This is the characteristic distractor design for sequence problems: the option set is populated with the sequence's own terms, so every reasonable amount of work lands on something that looks like an answer. Naming the position beside the number as you compute — 'third place, 10' rather than just '10' — removes the risk entirely.
• (d)5 — The FOURTH position's award, and also the value of the unknown a in the standard setting-out, which is what makes it so easy to select. Having written the four terms as a, a + 5, a + 10 and a + 15 and solved to a = 5, the temptation is to treat a as the answer; but a was defined as the smallest award, and the question asks for the third position, which is a + 5. It is worth noticing that 5 is also the common difference stated in the stem, so the same digit is doing two jobs on the page. Both errors are avoided by writing down explicitly what the unknown stands for before solving, and by returning to that definition before selecting.

AN ARITHMETIC PROGRESSION is a sequence in which each term differs from the previous one by a constant, the COMMON DIFFERENCE d. Its two working formulae are

nth term aₙ = a + (n − 1)d sum Sₙ = n/2 × [2a + (n − 1)d] = n/2 × (first term + last term)

The second form of the sum is the more useful one, because it says that the sum is the NUMBER OF TERMS multiplied by the AVERAGE OF THE FIRST AND LAST. And since an arithmetic progression is symmetric about its centre, that average is also the average of the whole sequence.

THE MEAN SHORTCUT follows directly and dissolves most examination problems of this type.

ODD number of terms — the mean IS the middle term. Three, five or seven terms summing to S give a middle term of S ÷ n at sight. EVEN number of terms — the mean lies exactly halfway between the two middle terms, so their sum is twice the mean.

That is why a four-term progression totalling 50 has its second and third terms summing to 25: no algebra is required to get there.

A SECOND DEVICE, symmetric labelling, is worth knowing for harder versions. Instead of writing four terms as a, a + d, a + 2d, a + 3d, write them as a − 3k, a − k, a + k, a + 3k with d = 2k. The sum is then simply 4a and the unknown difference vanishes from it. The same trick for three terms — a − d, a, a + d — makes the sum 3a. Problems that give a sum together with a product or a sum of squares become far easier this way.

WHAT MAKES THIS FAMILY OF PROBLEMS TRICKY is never the algebra. It is the bookkeeping: which end of the sequence is the largest, which position the question asks for, and what the unknown was defined to represent. In a race, in a prize list or in an age problem, the sequence has a direction imposed by the story, and the story has to be read before the formula is applied.

THE CHECK IS ALWAYS AVAILABLE. Any candidate value can be turned back into the full sequence and added up. A progression problem in which the total is given is self-verifying, and the verification takes less time than the original solution.

Key facts

• Four awards each 5 above the next form an arithmetic progression with common difference 5; writing the smallest as a gives 4a + 30 = 50, so a = 5.
• The full award is 20, 15, 10 and 5 for the first to fourth positions, and the third position therefore receives 10.
• The sum of an arithmetic progression is the number of terms times the average of the first and last: Sₙ = n/2 × (first + last).
• Because the terms are evenly spaced, the average of the whole progression is its centre — the middle term when the count is odd, and the midpoint of the two middle terms when it is even.
• Here 50 ÷ 4 = 12·5, so the second and third awards together are 25, which gives the third award as 10 without introducing an unknown.
• The nth term of an arithmetic progression is a + (n − 1)d.
• Symmetric labelling — a − 3k, a − k, a + k, a + 3k for four terms, or a − d, a, a + d for three — makes the common difference cancel out of the sum.
• Any candidate answer can be checked by rebuilding the whole sequence around it and adding: a third place of 30 would give 40, 35, 30, 25 = 130, not 50.

Common traps

• Answering with the value of the unknown you introduced rather than with the quantity asked for. Here a is the fourth place, not the third.
• Counting the positions from the wrong end, which turns the first place's award of 20 into an answer for the third.
• Reading a constant out of the working — the 30 in 4a + 30 = 50 is the pooled increments, not anyone's score.
• Getting the direction of the sequence wrong. A higher position takes more, so the run descends from first place to fourth.
• Reaching for the sum formula when the mean shortcut is faster: with the total given, the centre of the progression is already known.
• Skipping the rebuild-and-add check, which is quicker than the original solution and catches every one of these errors.

• practice — not a real PYQ

  Five prizes are to be awarded, each prize being ₹ 200 less than the one immediately above it, and the five together come to ₹ 5,000. What is the value of the third prize ?

  • (a)₹ 800
  • (b)₹ 1,000
  • (c)₹ 1,200
  • (d)₹ 1,400

  Answer(b) ₹ 1,000 — with an ODD number of terms in an arithmetic progression the average is the middle term itself, and the average here is 5,000 ÷ 5 = 1,000, which is the third prize. The full ladder is ₹ 1,400, ₹ 1,200, ₹ 1,000, ₹ 800 and ₹ 600, and those do sum to ₹ 5,000. No algebra is needed at all.

• practice — not a real PYQ

  Four numbers are in arithmetic progression with a common difference of 3, and their sum is 46. What is the smallest of the four numbers ?

  • (a)4
  • (b)7
  • (c)10
  • (d)13

  Answer(b) 7 — writing the smallest as a gives a + (a + 3) + (a + 6) + (a + 9) = 4a + 18 = 46, so a = 7 and the four numbers are 7, 10, 13 and 16. The other three options are the remaining terms of that same progression, which is exactly how sequence questions build their option sets: every value a solver passes through is offered alongside the one asked for.