In an arithmetic progression whose first term is (14) and fifteenth term is (84), what is the common difference?
Answer and explanation
Correct answer: 5
For an arithmetic progression, \(a_n=a+(n-1)d\). Here, \(a=14\) and \(a_{15}=84\), so \(84=14+(15-1)d=14+14d\). Thus, \(70=14d\), giving \(d=5\). If the common difference were 4, the fifteenth term would be \(14+14\times4=70\), not 84. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.
Frequently asked questions
What is the correct answer to this question?
5
Why is this the correct answer?
For an arithmetic progression, \(a_n=a+(n-1)d\). Here, \(a=14\) and \(a_{15}=84\), so \(84=14+(15-1)d=14+14d\). Thus, \(70=14d\), giving \(d=5\). If the common difference were 4, the fifteenth term would be \(14+14\times4=70\), not 84. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Arithmetic Progression.
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