In an AP (a_1+a_2=41) and (d=7). What is (a_{10})?
Answer and explanation
Correct answer: 80
Let the first term be \(a_1=a\). Then \(a_2=a+d=a+7\). Hence, \(a_1+a_2=a+(a+7)=41\), so \(2a=34\) and \(a=17\). Therefore, \(a_{10}=a+9d=17+9\times7=80\). Choosing 83 would result from using the term number incorrectly instead of \(n-1\). Exam tip: always use \(a_n=a+(n-1)d\) for the \(n\)th term of an AP.
Frequently asked questions
What is the correct answer to this question?
80
Why is this the correct answer?
Let the first term be \(a_1=a\). Then \(a_2=a+d=a+7\). Hence, \(a_1+a_2=a+(a+7)=41\), so \(2a=34\) and \(a=17\). Therefore, \(a_{10}=a+9d=17+9\times7=80\). Choosing 83 would result from using the term number incorrectly instead of \(n-1\). Exam tip: always use \(a_n=a+(n-1)d\) for the \(n\)th term of an AP.
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.
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