In an AP, (a_p=17), (a_{p+8}=65), and (p=6). What is (a_{30})?
Answer and explanation
Correct answer: 161
Since \(p=6\), we have \(a_p=a_6=17\) and \(a_{p+8}=a_{14}=65\). Thus, \(a_{14}-a_6=8d=65-17=48\), so \(d=6\). Now, \(a_{30}=a_6+(30-6)d=17+24\times6=161\). Hence, 161 is correct. Choosing 155 would not use the correct gap of \(30-6=24\) terms. Exam tip: In questions with indexed terms, substitute the value of \(p\) first and write the actual term numbers.
Frequently asked questions
What is the correct answer to this question?
161
Why is this the correct answer?
Since \(p=6\), we have \(a_p=a_6=17\) and \(a_{p+8}=a_{14}=65\). Thus, \(a_{14}-a_6=8d=65-17=48\), so \(d=6\). Now, \(a_{30}=a_6+(30-6)d=17+24\times6=161\). Hence, 161 is correct. Choosing 155 would not use the correct gap of \(30-6=24\) terms. Exam tip: In questions with indexed terms, substitute the value of \(p\) first and write the actual term numbers.
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.
