If (a_n=pn+q), (a_4+a_7=64), and (a_5+a_8=76), what will be (a_{10})?
Answer and explanation
Correct answer: 59
Given \(a_n=pn+q\), we have \(a_4+a_7=(4p+q)+(7p+q)=11p+2q=64\) and \(a_5+a_8=13p+2q=76\). Subtracting the first equation from the second gives \(2p=12\), so \(p=6\). Substituting into \(11p+2q=64\) gives \(q=-1\). Hence, \(a_{10}=10(6)-1=59\). The nearby option 61 can result from using an incorrect constant term. Exam tip: subtract paired-term equations first to eliminate \(q\) quickly.
Frequently asked questions
What is the correct answer to this question?
59
Why is this the correct answer?
Given \(a_n=pn+q\), we have \(a_4+a_7=(4p+q)+(7p+q)=11p+2q=64\) and \(a_5+a_8=13p+2q=76\). Subtracting the first equation from the second gives \(2p=12\), so \(p=6\). Substituting into \(11p+2q=64\) gives \(q=-1\). Hence, \(a_{10}=10(6)-1=59\). The nearby option 61 can result from using an incorrect constant term. Exam tip: subtract paired-term equations first to eliminate \(q\) quickly.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: nth term.
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