What is the sum of the first (21) even natural numbers?
Answer and explanation
Correct answer: (462)
The governing concept is the sum of an arithmetic progression. The first 21 even natural numbers are 2, 4, 6, …, 42, so they form an AP with a = 2, d = 2, n = 21, and last term l = 42. Using Sₙ = n(a + l) ÷ 2 gives S₂₁ = 21(2 + 42) ÷ 2 = 21 × 44 ÷ 2 = 21 × 22 = 462. Equivalently, the sum of the first n even numbers is n(n + 1), which gives 21 × 22 = 462. Therefore, option B is correct; the nearby alternatives reflect simple arithmetic errors.
Frequently asked questions
What is the correct answer to this question?
(462)
Why is this the correct answer?
The governing concept is the sum of an arithmetic progression. The first 21 even natural numbers are 2, 4, 6, …, 42, so they form an AP with a = 2, d = 2, n = 21, and last term l = 42. Using Sₙ = n(a + l) ÷ 2 gives S₂₁ = 21(2 + 42) ÷ 2 = 21 × 44 ÷ 2 = 21 × 22 = 462. Equivalently, the sum of the first n even numbers is n(n + 1), which gives 21 × 22 = 462. Therefore, option B is correct; the nearby alternatives reflect simple arithmetic errors.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.
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