In an arithmetic progression (a=11) and (d=6). If (S_n=1003), what is (n)?
Answer and explanation
Correct answer: 17
The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, 1003=\(\frac{n}{2}[22+6(n-1)]\)=\(n(3n+8)\). Hence, 3n²+8n−1003=0, giving n=17; the other root is negative. Although 19 is a nearby distractor, it does not give a sum of 1003. Exam tip: since n represents the number of terms, accept only a positive integer root.
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What is the correct answer to this question?
17
Why is this the correct answer?
The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, 1003=\(\frac{n}{2}[22+6(n-1)]\)=\(n(3n+8)\). Hence, 3n²+8n−1003=0, giving n=17; the other root is negative. Although 19 is a nearby distractor, it does not give a sum of 1003. Exam tip: since n represents the number of terms, accept only a positive integer root.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.
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