In an AP, \(a_3=11\) and \(a_{13}=51\). If \(a_{3r}=91\), what is (r)?
Answer and explanation
Correct answer: 7
In an AP, \(a_{13}-a_3=(13-3)d\). Hence \(51-11=10d\), so \(d=4\). Now use \(a_{3r}=a_3+(3r-3)d\): \(91=11+(3r-3)\times4\). Thus \(3r-3=20\), giving \(3r=21\) and \(r=7\). If \(r=6\), the term would be \(a_{18}\), whose value is 71, not 91. Exam tip: first find \(d\) from the two known terms and carefully use the difference between their indices.
Frequently asked questions
What is the correct answer to this question?
7
Why is this the correct answer?
In an AP, \(a_{13}-a_3=(13-3)d\). Hence \(51-11=10d\), so \(d=4\). Now use \(a_{3r}=a_3+(3r-3)d\): \(91=11+(3r-3)\times4\). Thus \(3r-3=20\), giving \(3r=21\) and \(r=7\). If \(r=6\), the term would be \(a_{18}\), whose value is 71, not 91. Exam tip: first find \(d\) from the two known terms and carefully use the difference between their indices.
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.
