SSC CGL 2024 Tier-I · 18 September 2024, Shift 3 · Quantitative Aptitude · Question 3 of 98

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 10 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

Find the average of first 125 natural numbers.

(a)63.5
(b)62
(c)63
(d)62.5

Answer

Why

Correct — C. The first 125 natural numbers are 1, 2, 3, …, 125, a run of consecutive integers.

Sum = n(n+1) ⁄ 2 = 125 × 126 ⁄ 2 = 7875
Average = 7875 ⁄ 125 = 63

Faster, skip the sum: for consecutive numbers the average is the middle term, and the middle of 125 terms is the 63rd, which is 63 → option (c)

Why the others are wrong

(a)
63.5 — 63.5 is 127 ⁄ 2 — the average of the first 126 natural numbers, one term too many. With an odd count of consecutive integers the mean lands on an actual term, so it cannot end in .5.
(b)
62 — 62 is the 62nd term, not the middle one. For 125 consecutive numbers the mean sits on the 63rd term, one place further along.
(d)
62.5 — 62.5 is n ⁄ 2 rather than (n+1) ⁄ 2. It is the correct average of 1 to 124, an even-length run whose mean falls between two terms.

Concept

For the first n natural numbers the average is (n + 1) ⁄ 2, and it is worth knowing why rather than memorising it.

The numbers form an arithmetic progression, and the mean of any AP is the mean of its first and last terms: (1 + 125) ⁄ 2 = 63.

That also explains the shape of the answer. An odd count of consecutive integers pairs off around a genuine middle term, so the mean is a whole number. An even count leaves the mean stranded halfway between two terms, ending in .5.

SSC counts the natural numbers from 1. Starting from 0 would make it a 126-term run from 0 to 125 with mean 62.5, which is where one of the wrong options comes from.

Key facts

The average of the first n natural numbers is (n + 1) ⁄ 2.
The sum of the first n natural numbers is n(n + 1) ⁄ 2, so 1 to 125 adds to 7875.
For any arithmetic progression the mean equals (first term + last term) ⁄ 2, here (1 + 125) ⁄ 2 = 63.
The average of the first n squares uses a different formula, (n + 1)(2n + 1) ⁄ 6.

Study next

Average of the first n natural numbers, of their squares, and of their cubesMean of an arithmetic progression as (first + last) ⁄ 2Average of consecutive odd or consecutive even numbers

Common traps

Using n ⁄ 2 and settling for 62.5.
Counting 0 as a natural number, which shifts the run and the mean.
Reaching for the squares formula when the question asks for the numbers themselves.

SSC asks this as one-line formula recall and then varies the count. The squares version turns up at 17 Sep 2024, 09:00, Quant Q.5 for the first 47 numbers and at 25 Sep 2024, 09:00, Quant Q.14 for the first 48. A set-defined cousin asks for the average of the primes between 50 and 76 at 11 Sep 2024, 12:30, Quant Q.18.

Related PYQs

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