The AP of multiples of (31) greater than (1200) is (1209,1240,1271,\ldots). What will be its (29)th term?
Answer and explanation
Correct answer: 2077
The first term of the AP is \(a=1209\), and the common difference is \(d=1240-1209=31\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{29}=1209+(29-1)\times31=1209+868=2077\). Hence, 2077 is correct. Using \(29\times31\) would incorrectly add 29 gaps; from the first term to the 29th term, there are only 28 gaps. Exam tip: always use \((n-1)\) in the nth-term formula.
Frequently asked questions
What is the correct answer to this question?
2077
Why is this the correct answer?
The first term of the AP is \(a=1209\), and the common difference is \(d=1240-1209=31\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{29}=1209+(29-1)\times31=1209+868=2077\). Hence, 2077 is correct. Using \(29\times31\) would incorrectly add 29 gaps; from the first term to the 29th term, there are only 28 gaps. Exam tip: always use \((n-1)\) in the nth-term formula.
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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