In the AP (31,45,59,\ldots), what is the greatest term between (900) and (1000)?
Answer and explanation
Correct answer: \(997\)
Here, the first term is \(a=31\) and the common difference is \(d=14\). Thus, \(a_n=31+14(n-1)\). To find the greatest term below \(1000\), use \(31+14(n-1)<1000\), which gives \(n-1\leq69\). Therefore, the term is \(31+14\times69=997\). The next term is \(1011\), which exceeds \(1000\). Exam tip: for the greatest AP term below a limit, use an inequality and take the greatest possible integer value of \(n-1\).
Frequently asked questions
What is the correct answer to this question?
\(997\)
Why is this the correct answer?
Here, the first term is \(a=31\) and the common difference is \(d=14\). Thus, \(a_n=31+14(n-1)\). To find the greatest term below \(1000\), use \(31+14(n-1)<1000\), which gives \(n-1\leq69\). Therefore, the term is \(31+14\times69=997\). The next term is \(1011\), which exceeds \(1000\). Exam tip: for the greatest AP term below a limit, use an inequality and take the greatest possible integer value of \(n-1\).
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.
