What will be the last term in the AP of positive multiples of (37) less than (3600)?
Answer and explanation
Correct answer: 3589
The sequence of positive multiples is \(37,74,111,\ldots\), whose \(n\)th term is \(37n\). For the last term, \(37n<3600\). Now \(37\times97=3589\), whereas \(37\times98=3626>3600\). Hence, 3589 is the greatest multiple of 37 less than 3600. Although 3626 is the next multiple, it exceeds the given limit. Exam tip: divide by the common difference, take the integer part, and multiply back to find the required multiple.
Frequently asked questions
What is the correct answer to this question?
3589
Why is this the correct answer?
The sequence of positive multiples is \(37,74,111,\ldots\), whose \(n\)th term is \(37n\). For the last term, \(37n<3600\). Now \(37\times97=3589\), whereas \(37\times98=3626>3600\). Hence, 3589 is the greatest multiple of 37 less than 3600. Although 3626 is the next multiple, it exceeds the given limit. Exam tip: divide by the common difference, take the integer part, and multiply back to find the required multiple.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the $n$th term of an AP.
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