What is the ratio of the sums of the first (96) and first (72) natural numbers?
Answer and explanation
Correct answer: 194:109
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{96}=\frac{96\times97}{2}=4656\) and \(S_{72}=\frac{72\times73}{2}=2628\). Dividing \(4656:2628\) by 24 gives \(194:109\). The ratio \(96:72\) compares only the numbers of terms, not the sums. Exam tip: when comparing such sums, include both factors \(n\) and \(n+1\).
Frequently asked questions
What is the correct answer to this question?
194:109
Why is this the correct answer?
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{96}=\frac{96\times97}{2}=4656\) and \(S_{72}=\frac{72\times73}{2}=2628\). Dividing \(4656:2628\) by 24 gives \(194:109\). The ratio \(96:72\) compares only the numbers of terms, not the sums. Exam tip: when comparing such sums, include both factors \(n\) and \(n+1\).
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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