Assistant Provident Fund Commissioner (APFC) 2016 recruitment test, advertisement 170. Transcribed from the Commission's own booklet (series A, B-CRHE-O-PAPA); answer as per the official UPSC final answer key. This question is in Part B. Answer last reviewed 24 August 2026.

Numbers a1, a2, a3, a4, a5, ..., a24 are in arithmetic progression and a1 + a5 + a10 + a15 + a20 + a24 = 225. The value of a1 + a2 + a3 + a4 + a5 + ... a23 + a24 is

(a)525
(b)725
(c)850
(d)900

Answer

Why

Correct — D, (d) 900.

The whole item turns on one property of an arithmetic progression: the sum of two terms depends only on the SUM OF THEIR POSITIONS, not on which two terms they are. With first term a1 and common difference d,

ap + aq = 2 x a1 + (p + q - 2) x d

so any two pairs of indices adding to the same total give the same sum.

Now look at the six indices the paper has chosen — 1, 5, 10, 15, 20, 24 — and pair them from the outside in:

1 + 24 = 25 5 + 20 = 25 10 + 15 = 25

All three pairs total 25, so all three pairs have the SAME sum, and that common sum is a1 + a24. The given equation is therefore

3 x (a1 + a24) = 225, so a1 + a24 = 75.

The sum of an arithmetic progression is the number of terms times the average of the first and last:

S24 = (24/2) x (a1 + a24) = 12 x 75 = 900.

The value of a1 + a2 + a3 + a4 + a5 + ... a23 + a24 is 900, which is option (d).

CHECK BY THE MEAN. The six given terms are symmetric about the centre of the progression, so their average is the average of the whole progression: 225/6 = 37·5. Twenty-four terms at an average of 37·5 give 24 x 37·5 = 900. The two routes agree, and the second is fast enough to use as a verification on any item of this shape.

Notice what the question does NOT require. Neither a1 nor d can be determined from the single equation given, and neither is needed. A candidate who sets up 2a1 + 23d and starts hunting for two unknowns from one equation will conclude the question is unanswerable. It is answerable precisely because the quantity asked for, like the quantity given, depends only on the combination a1 + a24.

Why the others are wrong

(a)525 — This does not come out of the data by any correct route, and the fastest way to see it is the average test. A total of 525 over 24 terms means an average term of 525/24, about 21·9. But the six given terms are symmetric about the centre of the progression and average 225/6 = 37·5, which must also be the average of the whole progression. Since 21·9 is not 37·5, the option cannot be right. It is the lowest of the four and serves as an anchor for a candidate who has decided the total should be a modest multiple of the given 225.
(b)725 — Also unreachable from the givens, and eliminated by the same one-line test: 725 over 24 terms is an average of about 30·2, against the required 37·5. It is placed between the two nearest wrong options to make the set look like a graded numerical scale, which is a common way of building an option list when only one value is derivable. Where three of four options cannot be produced by any route, the question is testing whether the structural insight was found at all, not the arithmetic that follows it.
(c)850 — The nearest miss, and the one most likely to be chosen by a candidate who has found the pairing but slipped in the final multiplication. It fails the average test too: 850/24 is about 35·4, not 37·5. It is worth noticing that all three wrong options lie BELOW 900. Every error available in this item — losing a pair, using six terms where twelve pairs are needed, or multiplying 75 by fewer than 12 — undercounts, so a candidate whose working has gone astray here will tend to land low rather than high.

Concept

An arithmetic progression is defined by a first term a1 and a constant common difference d, so that

an = a1 + (n - 1) x d.

Two consequences carry almost all AP questions.

THE PAIRING PROPERTY. ap + aq = 2a1 + (p + q - 2)d depends only on p + q. So in a progression of n terms,

a1 + an = a2 + a(n-1) = a3 + a(n-2) = ...

every pair equidistant from the two ends has the same sum. This is exactly why the sum formula works: pairing the n terms from the outside in gives n/2 pairs of equal value.

THE SUM. S(n) = (n/2) x (a1 + an) = (n/2) x [2a1 + (n-1)d]. The first form is the useful one whenever a1 + an can be got at directly, as here. Equivalently S(n) = n x (mean of the progression), and the mean of an AP is the average of its first and last terms — and also the middle term when n is odd.

THE DIAGNOSTIC that makes items like this quick: when a question gives you a sum of scattered terms, ADD THE INDICES IN PAIRS from the outside in. If they come out equal, the given sum is a multiple of a1 + an and the whole progression's total follows immediately. Here 1+24, 5+20 and 10+15 all give 25, which is the examiner telling you the item is solvable. If the index pairs did NOT match, no amount of algebra would extract the total from one equation in two unknowns.

The same symmetry underlies several other results worth carrying: the average of any set of terms symmetric about the centre equals the average of the whole progression; the sum of the first n natural numbers, n(n+1)/2, is the special case a1 = d = 1; and in a finite AP the mean, the median and the average of the extremes all coincide.

Sequences are a recurring form in the quantitative strand of this paper, which also sets a Fibonacci-style construction and several averaging items. This one is unusual in giving an equation that cannot be solved for its unknowns and asking for a quantity that nonetheless can be found — a deliberate test of whether the candidate reaches for structure or for algebra.

Two features of the printed page matter. Every index is a SUBSCRIPT in the booklet — a with subscript 1 through a with subscript 24 — and is written inline here as a1 through a24. And the two summation lines are set on their own lines, centred beneath the running text, which is where the line breaks in the transcribed stem come from.

There is also a misprint, reproduced as printed and present in both columns of the booklet. The second expression reads 'a1 + a2 + a3 + a4 + a5 + ... a23 + a24' — a plus sign BEFORE the ellipsis but none between the ellipsis and a23 — while the first expression prints its plus signs consistently. Nothing about the mathematics is in doubt: the expression is plainly the sum of all twenty-four terms, the stem says so by listing a1 through a5 and then a23 and a24, and the missing sign is a typesetting slip rather than an alternative reading. It has not been repaired here, and a candidate should read past it rather than around it.

Key facts

In an arithmetic progression, ap + aq = 2a1 + (p+q-2)d, so the sum of two terms depends only on the sum of their positions.
Terms equidistant from the two ends of a finite AP have equal pair sums: a1 + an = a2 + a(n-1) = a3 + a(n-2), and so on.
The six indices given here — 1, 5, 10, 15, 20, 24 — pair as 1+24, 5+20 and 10+15, each totalling 25, so the given sum is 3 x (a1 + a24).
a1 + a24 = 225/3 = 75, and S24 = (24/2) x 75 = 900.
Sum of an AP: S(n) = (n/2) x (a1 + an) = (n/2) x [2a1 + (n-1)d]; equivalently S(n) = n x (mean of the progression).
The mean check: any set of terms symmetric about the centre has the progression's own average, so 225/6 = 37·5 and 24 x 37·5 = 900.
Neither a1 nor d is determinable from the single equation given, and neither is needed — both the datum and the answer depend only on a1 + a24.

Study next

Arithmetic progressions: the nth term, the sum of n terms, and the pairing property that connects them.Arithmetic, geometric and harmonic means, and the standard inequality between them.Geometric progressions and the sum of a finite and an infinite series, which are asked in the same one-equation-two-unknowns style.Recognising when a question supplies a solvable COMBINATION of unknowns rather than the unknowns themselves.Symmetry arguments in averaging problems, including the average of consecutive integers and of terms picked symmetrically from a sequence.

Common traps

Trying to solve for a1 and d from one equation. Neither is determinable, and neither is required.
Missing that the six chosen indices pair to a constant total. Without that observation the given equation looks useless.
Using the number of TERMS where the number of PAIRS is needed, or the reverse. Twenty-four terms make twelve pairs of 75.
Skipping the mean check. Dividing each option by 24 and comparing with 225/6 = 37·5 eliminates the three wrong options in one line.

Sequence and series items on EPFO papers reward the structural observation rather than the manipulation. The standard construction gives partial information — a sum of scattered terms, a relation between two terms, a condition on the middle term — and asks for a total or for a particular term, and the information is always exactly enough for the quantity asked and never enough to pin down the progression. When you meet one, look first at the INDICES: add them in pairs from the outside in, count how many terms the answer spans, and check whether the set given is symmetric about the centre. Finish with the average test, which costs one division and catches an arithmetic slip that would otherwise be invisible.

Related PYQs

EPFO_APFC_2016_Q99

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

(a) 61·5
(b) 61
(c) 60·5
(d) 60

Answer(d) 60

Averages over a run of consecutive integers, 27 to 93 — the same symmetry argument, since consecutive integers are an arithmetic progression with common difference 1.

EPFO_APFC_2016_Q52

A certain sequence of integers is constructed as follows : Consider 0 and 1 as the first two numbers. The next, i.e. the third number is constructed by their sums, i.e. 1. This process of constructing the next number by the sum of the last two constructed numbers continues. Taking these numbers 0, 1, 1 as the first, second and third numbers in the sequence, what will be the 7th and 10th numbers, respectively ?

(a) 6 and 30
(b) 7 and 33
(c) 8 and 34
(d) 10 and 39

Answer(c) 8 and 34

The paper's other sequence-construction item, where each term is the sum of the previous two; both reward reading the rule of formation before computing any term.

EPFO_APFC_2016_Q95

Four quantities are such that their arithmetic mean (A.M.) is the same as the A.M. of the first three quantities. The fourth quantity is

(a) Sum of the first three quantities
(b) A.M. of the first three quantities
(c) (Sum of the first three quantities)/4
(d) (Sum of the first three quantities)/2

Answer(b) A.M. of the first three quantities

An averaging item comparing the arithmetic mean of four quantities with that of the first three, which turns on the same relation between a mean and a total.

Practice

practice — not a real PYQ
In an arithmetic progression of 20 terms, a3 + a18 = 40. What is the sum of all 20 terms ?
- (a)200
- (b)400
- (c)600
- (d)800
Answer(b) 400 — the indices 3 and 18 add to 21, and so do 1 and 20, so a3 + a18 = a1 + a20 = 40. The sum is (20/2) x 40 = 400. Neither the first term nor the common difference can be found, and neither is needed.
practice — not a real PYQ
If the sum of the first 15 terms of an arithmetic progression is 600, the eighth term is
- (a)30
- (b)40
- (c)45
- (d)75
Answer(b) 40 — with an odd number of terms the middle term is the mean of the progression, and the eighth of fifteen is the middle one. So a8 = 600/15 = 40. The same symmetry that pairs terms about the centre puts the mean at the centre when the count is odd.

Why