The sum of the first (n) natural numbers is (4656). What is the value of (n-10)?
Answer and explanation
Correct answer: 86
The sum of the first n natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=4656\), so \(n(n+1)=9312\). Since \(96\times97=9312\), \(n=96\). Therefore, \(n-10=96-10=86\). Choosing 84 would mean \(n=94\), whose sum is not 4656. Exam tip: when a sum is given, first find n using \(n(n+1)/2\), then evaluate the expression asked.
Frequently asked questions
What is the correct answer to this question?
86
Why is this the correct answer?
The sum of the first n natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=4656\), so \(n(n+1)=9312\). Since \(96\times97=9312\), \(n=96\). Therefore, \(n-10=96-10=86\). Choosing 84 would mean \(n=94\), whose sum is not 4656. Exam tip: when a sum is given, first find n using \(n(n+1)/2\), then evaluate the expression asked.
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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