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If (a_n=rn+s), (a_6=32), and (a_{11}=57), what will be (a_{20})?
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Answer and explanation
Correct answer: 102
Given \(a_n=rn+s\), we have \(a_{11}-a_6=5r\). Thus, \(57-32=25=5r\), so \(r=5\). Using \(a_6=6(5)+s=32\), we get \(s=2\). Hence, \(a_{20}=20(5)+2=102\). Option 100 ignores the constant term \(s\). Exam tip: subtracting two terms eliminates \(s\) and helps find \(r\) quickly.
Frequently asked questions
What is the correct answer to this question?
102
Why is this the correct answer?
Given \(a_n=rn+s\), we have \(a_{11}-a_6=5r\). Thus, \(57-32=25=5r\), so \(r=5\). Using \(a_6=6(5)+s=32\), we get \(s=2\). Hence, \(a_{20}=20(5)+2=102\). Option 100 ignores the constant term \(s\). Exam tip: subtracting two terms eliminates \(s\) and helps find \(r\) 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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