Consider the sequential integers 27 to 93, both included in the sequence. The arithmetic average of these numbers will be

Correct — D, (d) 60.

THE STEP THAT DECIDES THIS QUESTION is knowing that a run of consecutive integers is perfectly SYMMETRIC about its centre, so its average is simply the midpoint of its two ends:

average = (first + last) / 2 = (27 + 93) / 2 = 120 / 2 = 60

That is the whole calculation. No sum is needed and the number of terms never enters.

WHY THAT WORKS. Pair the run inwards from both ends: 27 with 93, 28 with 92, 29 with 91, and so on. Every pair adds to 120, and every pair therefore has an average of 60. A set built entirely out of pairs averaging 60 must itself average 60. If the count is odd there is one term left in the middle with no partner, and that middle term is 60 itself, so it changes nothing. The same argument works for any arithmetic progression, not just for consecutive integers, which is why the mean of an AP is always the average of its first and last terms whatever its length.

THE PARITY SHORTCUT, which disposes of half the option list before any arithmetic. Count the terms first:

number of terms = 93 − 27 + 1 = 67

Sixty-seven is ODD. A run with an odd number of terms has a genuine middle term, and for consecutive integers that middle term — and hence the average — is a WHOLE NUMBER. Half-integer averages ending in ·5 arise only when the count is EVEN, because then the centre falls between two integers. So options (a) and (c), 61·5 and 60·5, are impossible here on parity grounds alone and can be struck out in a few seconds. That leaves 61 and 60, and the midpoint calculation picks 60.

THE LONG ROUTE, as a check. The number of terms is 67 and the sum of an arithmetic progression is the number of terms times the average of the ends:

sum = 67 × 60 = 4020

Verify that independently. The sum of the integers from 1 to 93 is 93 × 94 / 2 = 4371, and the sum from 1 to 26 is 26 × 27 / 2 = 351, so the sum from 27 to 93 is 4371 − 351 = 4020. Dividing by 67 gives 60 again.

THE MIDDLE-TERM ROUTE, which is the fastest of all once the count is known. With 67 terms the middle one is the thirty-fourth, and the thirty-fourth term of a run beginning at 27 is 27 + 33 = 60. For consecutive integers the mean and the median coincide exactly, because the distribution is symmetric, so finding the middle term and finding the average are the same operation.

Three independent routes, all landing on 60, which is option (d). The decimals in two of the options are printed with a raised middle dot rather than a full stop, as this booklet does throughout.

Why the others are wrong

• (a)61·5 — This is the largest option and it is ruled out twice over. First by parity: the run from 27 to 93 contains 93 − 27 + 1 = 67 terms, an odd number, so its average is the middle term and must be a whole number — no run of consecutive integers with an odd count can average a half. Second by position: 61·5 is a full one and a half above the midpoint of the two ends, and the average of a symmetric run cannot lie away from its centre at all. A half-integer of this size belongs to a different range: a run from 28 to 95, or from 30 to 93, has an even number of terms and averages 61·5. The general safeguard is to compute the count before the average and read its parity, which decides in advance whether the answer can carry a fraction. It is one of the cheapest checks in arithmetic and it eliminates half of this option list.
• (b)61 — Sixty-one is one more than the average and is very likely the product of a fencepost error — the commonest single mistake in problems about runs of numbers. The sum of the run is genuinely 4020, and a candidate who obtains that correctly but then divides by 93 − 27 = 66 instead of by 67 gets 60·909…, which rounds to 61. The subtraction 93 − 27 counts the GAPS between the numbers, not the numbers themselves, and there is always one more number than there are gaps: from 27 to 93 inclusive there are 66 steps and 67 integers. The same slip in reverse — using 68 — would give 59·1. The defence is to test the formula on a tiny case whenever it is used: the run from 3 to 5 obviously contains three numbers, and 5 − 3 + 1 = 3 confirms that the ordinary subtraction alone would have said two. Note also that this option cannot be reached by any correct method, since the true average is exactly the midpoint of 27 and 93.
• (c)60·5 — Sixty and a half sits half a unit above the answer and falls to the same parity argument as the other decimal option. The run holds 67 numbers, an odd count, so its centre is an actual member of the run and the average is a whole number. A half-integer average would require an even count, which is what happens if either endpoint is moved by one: the run from 27 to 94 contains 68 numbers and averages exactly 60·5. That is a useful way to read this option — it is the answer to the question with one extra number added at the top, and a candidate who mis-transcribes the upper bound, or who half-remembers that something must be added because of the inclusive endpoints, lands here. The correction is to keep the two operations separate. The endpoints affect the COUNT through the plus one; they affect the AVERAGE only through their own values, and (27 + 93)/2 uses those values directly with no adjustment of any kind.

THE ARITHMETIC PROGRESSION AND ITS AVERAGE. A run of consecutive integers is the simplest arithmetic progression there is: each term exceeds the one before by a common difference of 1. Every AP has three facts attached to it that between them answer almost any question a paper can ask.

Number of terms: n = (last − first) / d + 1, which for consecutive integers is last − first + 1 Average: (first + last) / 2 Sum: n × (first + last) / 2, that is, the number of terms times the average

The middle one is the fact this question needs, and it is worth understanding rather than memorising. An AP is symmetric about its centre: terms equidistant from the two ends are equidistant from the middle in opposite directions, so their deviations cancel. Since the arithmetic mean is the point about which deviations cancel, the mean must be that centre — and the centre of a symmetric run is the average of its extremes. Notice that the LENGTH of the run is irrelevant to its average. A run from 27 to 93 and a much sparser progression from 27 to 93 in steps of 3 have the same average of 60.

MEAN AND MEDIAN COINCIDE for any symmetric distribution, and an AP is exactly symmetric. That gives a second route to the answer: find the middle term. With an odd count the middle term is a real member of the set; with an even count the median is the average of the two central terms and is a half-integer whenever the terms are integers. This is why the parity of the count controls whether the answer can carry a fraction.

THE FENCEPOST RULE deserves separate billing because it accounts for so many wrong answers in this area. Between the first and last of n objects laid in a line there are n − 1 gaps, so counting objects by subtracting their labels always undercounts by one. From 27 to 93 there are 66 steps and 67 numbers. The rule turns up everywhere — in the number of terms of a series, in the number of pages of a book, in the number of days between two dates, in the number of posts in a fence — and the safest habit is to test any counting formula on a case small enough to check by eye.

FINALLY, THE PARITY CHECK AS A TOOL. Before computing an average of integers, ask how many there are. An odd count of consecutive integers gives a whole-number average; an even count gives one ending in a half. On a multiple-choice paper that single observation frequently halves the option list, and it costs one subtraction.

Key facts

• The average of any arithmetic progression is the average of its first and last terms, so for the integers from 27 to 93 it is (27 + 93)/2 = 60.
• The run contains 93 − 27 + 1 = 67 integers; subtracting alone counts the 66 gaps between them rather than the numbers themselves.
• Because 67 is odd, the run has a genuine middle term and its average must be a whole number, which rules out both half-integer options at once.
• The middle term is the thirty-fourth, and 27 + 33 = 60, so the mean and the median of the run are the same number.
• The sum is 67 × 60 = 4020, which checks against 4371 − 351, the sum from 1 to 93 less the sum from 1 to 26.
• For any symmetric distribution the mean equals the median, and an arithmetic progression is exactly symmetric about its centre.
• The length of a progression does not affect its average; only its two extreme terms do, which is why the count is never needed here.
• A run of consecutive integers averages a half-integer only when it contains an even number of terms — from 27 to 94, for instance, the average is 60·5.

Common traps

• Dividing the sum by the difference of the endpoints rather than by the count. From 27 to 93 there are 66 gaps but 67 numbers.
• Overlooking the parity check. An odd count of consecutive integers cannot average a half, which removes two options immediately.
• Computing the sum when only the average is wanted, which adds two chances of arithmetic error and answers a question nobody asked.
• Applying the plus one to the average as well as to the count. The endpoints enter the average directly through (first + last)/2.
• Assuming the number of terms is needed for the average. It is not; the average of a progression depends only on its two ends.

• practice — not a real PYQ

  What is the arithmetic average of all the integers from 41 to 79, taking both of those numbers as included in the run ?

  • (a)59
  • (b)60
  • (c)61
  • (d)62

  Answer(b) 60 — the run is symmetric about its centre, so the average is the midpoint of the two ends, (41 + 79)/2 = 60. The run holds 79 − 41 + 1 = 39 integers, an odd count, which confirms that the average must be a whole number.

• practice — not a real PYQ

  In an arithmetic progression the first term is 7 and the last term is 97. What is the arithmetic average of all the terms of that progression ?

  • (a)48
  • (b)52
  • (c)56
  • (d)It cannot be found unless the number of terms is known

  Answer(b) 52 — the average of any arithmetic progression is the average of its extreme terms, (7 + 97)/2 = 52, and it does not depend on how many terms lie between them. That is why the last option, which looks cautious, is wrong.