PUNE MCA Exam 2016 Question Paper with answers for online practice

Concept:The letters are arranged in a triangular pattern starting from the top with A, each subsequent row having one more letter. The leftmost letter of the $n$‑th row is the $(1 + \frac{n(n-1)}{2})$‑th letter of the alphabet.Explanation:Assuming a common pyramid arrangement: row 1: A; row 2: B, C; row 3: D, E, F; etc.The number of letters before the $n$‑th row is $\frac{(n-1)n}{2}$, so the first letter of the $n$‑th row is the $\left(\frac{n(n-1)}{2}+1\right)$‑th letter.For the lowest row, we need a result that matches one of the given options. Only $n=5$ gives $\frac{5 \times 4}{2}+1 = 11$, which is the 11th letter, K.Thus the leftmost letter in the lowest row is K.Answer:Option A: K