Solution (AP method):

Total children = 630. Let rows = n, first row = a, common difference d = -3.

Sum of n terms: \(S_n=\frac{n}{2}[2a+(n-1)d]=630\)

\(\Rightarrow 1260 = n\,[2a-3(n-1)] \;\Rightarrow\; 2a=\frac{1260}{n}+3(n-1)\)

\(\Rightarrow a=\frac{1}{2}\left(\frac{1260}{n}+3n-3\right)\). For a valid arrangement, a must be a positive integer and last row \(a-3(n-1)>0\).

• n = 3: \(a=\tfrac{1}{2}(420+9-3)=\tfrac{1}{2}\cdot 426=213\) ✔️ integer (valid)
• n = 4: \(a=\tfrac{1}{2}(315+12-3)=\tfrac{1}{2}\cdot 324=162\) ✔️ integer (valid)
• n = 5: \(a=\tfrac{1}{2}(252+15-3)=\tfrac{1}{2}\cdot 264=132\) ✔️ integer (valid)
• n = 6: \(a=\tfrac{1}{2}(210+18-3)=\tfrac{1}{2}\cdot 225=112.5\) ❌ not integer (invalid)

✅ Final Answer: 6 rows is not possible.