GATE 2026 EC GA Q.8: solution and answer

GATE 2026 · ECGA Q.8 · 2 marks · MCQ

Let $P_{k}$ represent the perimeter of a square with sides of length $k$ . The value of the expression $P_{1} + P_{2} + P_{3} + \dots + P_{10}$ is ______

(A)
55
(B)
110
(C)
220
(D)
440

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Answer and solution

ANSWER

$P_{1} + P_{2} + \dots + P_{10} = 4 (1 + 2 + \dots + 10) = 4 \times 55 = 220$ , option (C).

Official answer: CSource: Official answer key, GATE 2026 EC (IIT Guwahati)

A square has 4 equal sides, so the perimeter of a square of side $k$ is $P_{k} = 4 k$ .
So the sum is $P_{1} + P_{2} + \dots + P_{10} = 4 (1 + 2 + \dots + 10)$ . Take the common factor 4 out first.
The sum of the first $n$ natural numbers is $\frac{n ( n + 1 )}{2}$ . With $n = 10$ : $1 + 2 + \dots + 10 = \frac{10 \times 11}{2} = 55$ .
$P_{1} + P_{2} + \dots + P_{10} = 4 \times 55 = 220$
Check by adding the terms directly: $4 + 8 + 12 + 16 + 20 + 24 + 28 + 32 + 36 + 40 = 220$ . So the answer is 220, option (C).

KEY IDEA

Write each term in terms of $k$ ( $P_{k} = 4 k$ ), take the common factor 4 out, and use $1 + 2 + \dots + n = \frac{n ( n + 1 )}{2}$ .

WATCH OUT FOR

Summing the side lengths and stopping there gives $1 + 2 + \dots + 10 = 55$ (option A). Using $2 k$ per square, i.e. only two sides or a semi-perimeter, gives 110 (option B). Using the square's area-like factor or doubling the perimeter by mistake gives 440 (option D).

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GATE 2026 EC GA Q.8: solution and answer

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