What will be the last term in the AP of positive multiples of (29) less than (2500)?
Answer and explanation
Correct answer: 2494
The positive multiples of 29 form the AP \(29, 58, 87, \ldots\), whose \(n\)th term is \(29n\). For the last term, \(29n<2500\). Since \(2500\div29\approx86.2\), the greatest integer value of \(n\) is 86. Hence the last term is \(29\times86=2494\). Although \(2523=29\times87\), it is greater than 2500. Exam tip: for “less than”, use the integer part of the quotient and verify the product.
Frequently asked questions
What is the correct answer to this question?
2494
Why is this the correct answer?
The positive multiples of 29 form the AP \(29, 58, 87, \ldots\), whose \(n\)th term is \(29n\). For the last term, \(29n<2500\). Since \(2500\div29\approx86.2\), the greatest integer value of \(n\) is 86. Hence the last term is \(29\times86=2494\). Although \(2523=29\times87\), it is greater than 2500. Exam tip: for “less than”, use the integer part of the quotient and verify the product.
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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