Question

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

• A. 3
• B. 4
• C. 5
• D. 6
• E. 7

Hint:

Check feasibility by ensuring the first term $a$ is a positive integer when using the arithmetic progression sum formula.

Answer 1

The Correct Option is D

Let $n$ = number of rows, $a$ = children in first row, common difference $d = -3$. Total children: \[ S_n = \frac{n}{2} [2a + (n-1)d] = 630 \] For $n=6$: \[ \frac{6}{2}[2a + 5(-3)] = 3(2a - 15) = 630 \Rightarrow 2a - 15 = 210 \Rightarrow 2a = 225 \Rightarrow a = 112.5 \] Non-integer → Not possible. For other $n$, $a$ is integer. \[ \boxed{6} \]