What is the value of (S_{18}+S_{12}), where (S_n) is the sum of the first (n) natural numbers?
Answer and explanation
Correct answer: 249
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{18}=\frac{18\times19}{2}=171\) and \(S_{12}=\frac{12\times13}{2}=78\). Therefore, \(S_{18}+S_{12}=171+78=249\). Option 239 is incorrect because it is not the sum of these two correct values. Exam tip: In such questions, first use \(S_n=\frac{n(n+1)}{2}\) to find each required sum separately.
Frequently asked questions
What is the correct answer to this question?
249
Why is this the correct answer?
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{18}=\frac{18\times19}{2}=171\) and \(S_{12}=\frac{12\times13}{2}=78\). Therefore, \(S_{18}+S_{12}=171+78=249\). Option 239 is incorrect because it is not the sum of these two correct values. Exam tip: In such questions, first use \(S_n=\frac{n(n+1)}{2}\) to find each required sum separately.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Sum of first n natural numbers.
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