What is the sum of the first (16) even natural numbers?
Answer and explanation
Correct answer: 272
The first 16 even natural numbers are 2, 4, 6, ..., 32. They form an AP with \(a=2\), \(d=2\), and \(n=16\). Therefore, \(S_{16}=\frac{16}{2}[2(2)+(16-1)\cdot2]=272\). Hence, 272 is correct. Although 256 equals \(16^2\), the sum of the first 16 even natural numbers is given by \(n(n+1)\). Exam tip: use \(n(n+1)\) directly for the sum of the first \(n\) even natural numbers.
Frequently asked questions
What is the correct answer to this question?
272
Why is this the correct answer?
The first 16 even natural numbers are 2, 4, 6, ..., 32. They form an AP with \(a=2\), \(d=2\), and \(n=16\). Therefore, \(S_{16}=\frac{16}{2}[2(2)+(16-1)\cdot2]=272\). Hence, 272 is correct. Although 256 equals \(16^2\), the sum of the first 16 even natural numbers is given by \(n(n+1)\). Exam tip: use \(n(n+1)\) directly for the sum of the first \(n\) even natural numbers.
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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