The AP of multiples of (37) greater than (1500) is (1517,1554,1591,\ldots). What will be its (31)st term?
Answer and explanation
Correct answer: 2627
The first term of the AP is \(a=1517\), and its common difference is \(d=1554-1517=37\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Hence, \(a_{31}=1517+(31-1)\times37=1517+1110=2627\). Therefore, option B is correct. Using \(2609\) results from an incorrect term count or from not using \((n-1)\). Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.
Frequently asked questions
What is the correct answer to this question?
2627
Why is this the correct answer?
The first term of the AP is \(a=1517\), and its common difference is \(d=1554-1517=37\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Hence, \(a_{31}=1517+(31-1)\times37=1517+1110=2627\). Therefore, option B is correct. Using \(2609\) results from an incorrect term count or from not using \((n-1)\). Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.
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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