Question

A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

Apply sum of arithmetic progression formula and check each option — discard those that lead to non-integer terms.

Answer 1

The Correct Option is D

Let the number of rows be \( n \) and first row has \( a \) children. Each next row has 3 fewer children: This is an arithmetic series: \[ \text{Total children} = \frac{n}{2} [2a - (n-1) \cdot 3] = 630 \] We check for all given options to find which gives non-integral or invalid solution. \underline{Option (A): \(n = 3\)} \[ \frac{3}{2}[2a - 6] = 630 \Rightarrow 3(2a - 6) = 1260 \Rightarrow 2a - 6 = 420 \Rightarrow a = 213 \] \underline{Option (B): \(n = 4\)} \[ \frac{4}{2}[2a - 9] = 630 \Rightarrow 2(2a - 9) = 630 \Rightarrow 2a - 9 = 315 \Rightarrow a = 162 \] \underline{Option (C): \(n = 5\)} \[ \frac{5}{2}[2a - 12] = 630 \Rightarrow 5(2a - 12) = 1260 \Rightarrow 2a - 12 = 252 \Rightarrow a = 132 \] \underline{Option (D): \(n = 6\)} \[ \frac{6}{2}[2a - 15] = 630 \Rightarrow 3(2a - 15) = 630 \Rightarrow 2a - 15 = 210 \Rightarrow a = 112.5 \quad \text{Not possible} \] Since \(a\) is not an integer, Option (D) is not valid. \fbox{Final Answer: (D) 6}