The sum of the first (18) terms is (441), and the first term is (3). If the sequence is an AP, what is the common difference (d)?
Answer and explanation
Correct answer: \(\frac{43}{17}\)
For an AP, \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(441=\frac{18}{2}[2(3)+17d]=9(6+17d)\). Hence \(49=6+17d\), so \(17d=43\) and \(d=\frac{43}{17}\). If \(d=3\), the sum would be \(513\), so it is not correct. Exam tip: first divide by \(\frac{n}{2}\) to simplify the equation.
Frequently asked questions
What is the correct answer to this question?
\(\frac{43}{17}\)
Why is this the correct answer?
For an AP, \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(441=\frac{18}{2}[2(3)+17d]=9(6+17d)\). Hence \(49=6+17d\), so \(17d=43\) and \(d=\frac{43}{17}\). If \(d=3\), the sum would be \(513\), so it is not correct. Exam tip: first divide by \(\frac{n}{2}\) to simplify the equation.
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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