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 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?
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Check feasibility by ensuring the first term $a$ is a positive integer when using the arithmetic progression sum formula.
Updated On: Jul 31, 2025
- 3
- 4
- 5
- 6
- 7
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The Correct Option is D
Solution and Explanation
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.
\[
\boxed{6}
\]
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