What will be the last term in the AP of positive multiples of (41) less than (5000)?
Answer and explanation
Correct answer: 4961
The positive multiples form the AP \(41, 82, 123, \ldots\). The last term must be less than 5000. Since \(41\times121=4961\), and the next multiple is \(41\times122=5002\), which is greater than 5000, the last term is \(4961\). Also, \(4920\) is not a multiple of 41. Exam tip: divide by the given number, take the integer part, and verify the resulting multiple against the limit.
Frequently asked questions
What is the correct answer to this question?
4961
Why is this the correct answer?
The positive multiples form the AP \(41, 82, 123, \ldots\). The last term must be less than 5000. Since \(41\times121=4961\), and the next multiple is \(41\times122=5002\), which is greater than 5000, the last term is \(4961\). Also, \(4920\) is not a multiple of 41. Exam tip: divide by the given number, take the integer part, and verify the resulting multiple against the limit.
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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