In the AP (58,51,44,\ldots), which is the first term less than (-60)?
Answer and explanation
Correct answer: -61
The first term is \(a=58\) and the common difference is \(d=-7\). Hence, \(a_n=58+(n-1)(-7)=65-7n\). For \(a_n<-60\), we need \(65-7n<-60\), so \(n>125/7\). The smallest integer satisfying this is \(n=18\). Therefore, \(a_{18}=65-7(18)=-61\). The previous term is \(a_{17}=-54\), which is not less than \(-60\). Although \(-68\) is also less than \(-60\), it occurs later in the AP. Exam tip: when asked for the first term meeting an inequality, use the smallest integer \(n\) that satisfies it.
Frequently asked questions
What is the correct answer to this question?
-61
Why is this the correct answer?
The first term is \(a=58\) and the common difference is \(d=-7\). Hence, \(a_n=58+(n-1)(-7)=65-7n\). For \(a_n<-60\), we need \(65-7n<-60\), so \(n>125/7\). The smallest integer satisfying this is \(n=18\). Therefore, \(a_{18}=65-7(18)=-61\). The previous term is \(a_{17}=-54\), which is not less than \(-60\). Although \(-68\) is also less than \(-60\), it occurs later in the AP. Exam tip: when asked for the first term meeting an inequality, use the smallest integer \(n\) that satisfies it.
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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