If an arithmetic progression has first term \(a\), common difference \(d\), and \(n\) terms, which is the correct formula for the sum \(S_n\) of its first \(n\) terms?
Answer and explanation
Correct answer: \(S_n=\frac{n}{2}[2a+(n-1)d]\)
The nth, or last, term of the AP is \(l=a+(n-1)d\). Substituting this in \(S_n=\frac{n}{2}(a+l)\) gives \(S_n=\frac{n}{2}[2a+(n-1)d]\). Option B is only the formula for the nth term, not the sum, while option C incorrectly uses \(nd\) in place of \((n-1)d\). Exam tip: always check both the factor \(n\) and \((n-1)d\) in the sum formula.
Frequently asked questions
What is the correct answer to this question?
\(S_n=\frac{n}{2}[2a+(n-1)d]\)
Why is this the correct answer?
The nth, or last, term of the AP is \(l=a+(n-1)d\). Substituting this in \(S_n=\frac{n}{2}(a+l)\) gives \(S_n=\frac{n}{2}[2a+(n-1)d]\). Option B is only the formula for the nth term, not the sum, while option C incorrectly uses \(nd\) in place of \((n-1)d\). Exam tip: always check both the factor \(n\) and \((n-1)d\) in the sum formula.
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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