If \(a_n=an+b\), \(a_4=21\), and \(a_{14}=71\), what is \(a_{30}\)?
Answer and explanation
Correct answer: 151
Given \(a_n=an+b\), we have \(a_{14}-a_4=10a\). Thus, \(71-21=10a\), so \(a=5\). From the 14th term to the 30th term, there are 16 term-intervals. Therefore, \(a_{30}=a_{14}+16\times5=71+80=151\). The value 147 would result from using an incorrect interval count. Exam tip: subtract two given terms to eliminate the constant \(b\) and find the common difference/slope directly.
Frequently asked questions
What is the correct answer to this question?
151
Why is this the correct answer?
Given \(a_n=an+b\), we have \(a_{14}-a_4=10a\). Thus, \(71-21=10a\), so \(a=5\). From the 14th term to the 30th term, there are 16 term-intervals. Therefore, \(a_{30}=a_{14}+16\times5=71+80=151\). The value 147 would result from using an incorrect interval count. Exam tip: subtract two given terms to eliminate the constant \(b\) and find the common difference/slope directly.
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.
