The first term of an AP is (31) and (a_{19}=139). What is the common difference?
Answer and explanation
Correct answer: 6
For an AP, \(a_n=a+(n-1)d\). Substituting \(a=31\), \(n=19\), and \(a_{19}=139\), we get \(139=31+18d\). Thus, \(108=18d\), so \(d=6\). If the common difference were 5, the 19th term would be \(31+18\times5=121\), not 139. Exam tip: the common difference is added \(n-1\) times to reach the \(n\)th term.
Frequently asked questions
What is the correct answer to this question?
6
Why is this the correct answer?
For an AP, \(a_n=a+(n-1)d\). Substituting \(a=31\), \(n=19\), and \(a_{19}=139\), we get \(139=31+18d\). Thus, \(108=18d\), so \(d=6\). If the common difference were 5, the 19th term would be \(31+18\times5=121\), not 139. Exam tip: the common difference is added \(n-1\) times to reach the \(n\)th term.
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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