An arithmetic progression (AP) has first term \(a\) and common difference \(d\). Which formula correctly represents the sum \(S_n\) of its first \(n\) terms?
Answer and explanation
Correct answer: \(S_n=\frac{n}{2}[2a+(n-1)d]\)
The \(n\)th term of an AP is \(a+(n-1)d\). Therefore, the average of the first and last terms is \(\frac{a+[a+(n-1)d]}{2}\). Multiplying this by \(n\) gives \(S_n=\frac{n}{2}[2a+(n-1)d]\). Option B is only the \(n\)th term, while option C incorrectly uses \(nd\) instead of \((n-1)d\). Exam tip: always check the \((n-1)d\) factor for the last term.
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 \(n\)th term of an AP is \(a+(n-1)d\). Therefore, the average of the first and last terms is \(\frac{a+[a+(n-1)d]}{2}\). Multiplying this by \(n\) gives \(S_n=\frac{n}{2}[2a+(n-1)d]\). Option B is only the \(n\)th term, while option C incorrectly uses \(nd\) instead of \((n-1)d\). Exam tip: always check the \((n-1)d\) factor for the last term.
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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