In an AP, (a_1+a_2=33) and (d=5). What is (a_{12})?
Answer and explanation
Correct answer: 69
In an AP, \(a_1=a\) and \(a_2=a+d\). Hence, \(a_1+a_2=a+(a+5)=33\), so \(2a=28\) and \(a=14\). Therefore, \(a_{12}=a+11d=14+11\times5=69\). Choosing 71 would result from using the number of terms or the common difference incorrectly. Exam tip: while finding the \(n\)th term, use \(a_n=a+(n-1)d\), ensuring that the multiplier is \(n-1\).
Frequently asked questions
What is the correct answer to this question?
69
Why is this the correct answer?
In an AP, \(a_1=a\) and \(a_2=a+d\). Hence, \(a_1+a_2=a+(a+5)=33\), so \(2a=28\) and \(a=14\). Therefore, \(a_{12}=a+11d=14+11\times5=69\). Choosing 71 would result from using the number of terms or the common difference incorrectly. Exam tip: while finding the \(n\)th term, use \(a_n=a+(n-1)d\), ensuring that the multiplier is \(n-1\).
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.
