The first term of an AP is (x), and the common difference is (x+2). If the sum of the first (10) terms is (365), what is the value of (x)?

\(5\)

Here, the first term is \(a=x\), the common difference is \(d=x+2\), and \(n=10\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(365=\frac{10}{2}[2x+9(x+2)]\). Thus, \(365=5(11x+18)=55x+90\), so \(x=5\). Substituting \(x=6\) gives a sum of \(420\), so it is incorrect despite being the closest distractor. Exam tip: In AP questions involving variables, identify \(a\), \(d\), and \(n\) before applying the formula.

This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.