If \(a_n=rn+s\), \(a_3=14\), and \(a_{10}=49\), what will be \(a_{15}\)?
Answer and explanation
Correct answer: 74
Because \(a_n=rn+s\) is linear, the change in the term is proportional to the change in the index. Between the third and tenth terms, the value increases by \(49-14=35\) over \(10-3=7\) index intervals. Thus \(7r=35\), so \(r=5\). Using \(a_3=14\), we obtain \(3(5)+s=14\), hence \(s=-1\). Therefore \(a_{15}=5(15)-1=75-1=74\), so option B is correct. Equivalently, from the tenth to the fifteenth term there are five steps, giving \(49+5\cdot5=74\).
Frequently asked questions
What is the correct answer to this question?
74
Why is this the correct answer?
Because \(a_n=rn+s\) is linear, the change in the term is proportional to the change in the index. Between the third and tenth terms, the value increases by \(49-14=35\) over \(10-3=7\) index intervals. Thus \(7r=35\), so \(r=5\). Using \(a_3=14\), we obtain \(3(5)+s=14\), hence \(s=-1\). Therefore \(a_{15}=5(15)-1=75-1=74\), so option B is correct. Equivalently, from the tenth to the fifteenth term there are five steps, giving \(49+5\cdot5=74\).
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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