The sum of 18 consecutive natural numbers is a perfect square. What is the smallest possible value of this sum?

### Concept & Formula The sum of an arithmetic progression (AP) is calculated using the formula: $$S_n = \frac{n}{2} [2a + (n-1)d]$$ For consecutive natural numbers, the common difference $d = 1$. The sum must then be evaluated to find what conditions make it a perfect square. ### Step-by-Step Solution * **Given:** Number of terms $n = 18$. The sum $S_{18}$ is a perfect square. We need the smallest possible value. * **Calculation / Deduction:** * Let the $18$ consecutive natural numbers start from $a$. (Where $a \ge 1$) * Using the AP sum formula: $$S_{18} = \frac{18}{2} [2a + (18-1)(1)]$$ $$S_{18} = 9 [2a + 17]$$ * For $S_{18}$ to be a perfect square, the expression $9(2a + 17)$ must be a perfect square. * Since $9$ is already a perfect square ($3^2$), the factor $(2a + 17)$ must also be a perfect square. * Because $a$ is a natural number ($a \ge 1$), the minimum value of $(2a + 17)$ is $2(1) + 17 = 19$. * We need the smallest perfect square greater than or equal to $19$. * The perfect squares are $1, 4, 9, 16, 25, 36 \dots$ * The smallest perfect square greater than $19$ is $25$. * So, we set $2a + 17 = 25$. $$2a = 8 \Rightarrow a = 4$$ * The condition holds. Now calculate the sum: $$S_{18} = 9 \times 25 = 225$$ ### Exam Strategy & Shortcut You don't need to calculate $a$. Once you realize the sum is $9 \times (\text{an odd number greater than } 17)$, you know the odd number must be a perfect square. The first perfect square after $17$ is $25$. The minimum sum is simply $9 \times 25 = 225$. ### Common Pitfall A common error is forgetting that $a$ must be a natural number ($a \ge 1$), which sets a lower limit on $2a+17$. Students might just look for *any* perfect square multiplier, but checking the boundary condition ($2a+17 \ge 19$) is crucial to finding the true minimum sum. ### Final Answer **Therefore, the correct answer is 225.**