What is the last term in the AP of positive multiples of (23) less than (2000)?
Answer and explanation
Correct answer: 1978
The sequence of positive multiples is \(23, 46, 69, \ldots\), and its \(n\)th term is \(23n\). For the last term, \(23n<2000\). Since \(2000\div23\approx86.95\), the greatest integer value of \(n\) is 86. Thus, the last term is \(23\times86=1978\). Although \(2001=23\times87\), it is greater than 2000. Exam tip: for “less than” questions, check that the next multiple does not cross the given limit.
Frequently asked questions
What is the correct answer to this question?
1978
Why is this the correct answer?
The sequence of positive multiples is \(23, 46, 69, \ldots\), and its \(n\)th term is \(23n\). For the last term, \(23n<2000\). Since \(2000\div23\approx86.95\), the greatest integer value of \(n\) is 86. Thus, the last term is \(23\times86=1978\). Although \(2001=23\times87\), it is greater than 2000. Exam tip: for “less than” questions, check that the next multiple does not cross the given 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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