The sum of how many terms of the $A.P.$ $54, 51, 48, \dots$ is $513$? Explain the reason for getting two different answers.

(A) For the $A.P.$ $54, 51, 48, \dots$,the first term $a = 54$,the common difference $d = 51 - 54 = -3$,and the sum of $n$ terms $S_{n} = 513$.
Using the formula $S_{n} = \frac{n}{2}[2a + (n - 1)d]$:
$513 = \frac{n}{2}[2(54) + (n - 1)(-3)]$
$1026 = n[108 - 3n + 3]$
$1026 = n[111 - 3n]$
$1026 = 111n - 3n^{2}$
$3n^{2} - 111n + 1026 = 0$
Dividing by $3$:
$n^{2} - 37n + 342 = 0$
Factoring the quadratic equation:
$(n - 18)(n - 19) = 0$
Thus,$n = 18$ or $n = 19$.
Reason for two answers: The $19^{th}$ term of the $A.P.$ is $T_{19} = a + 18d = 54 + 18(-3) = 54 - 54 = 0$. Since the $19^{th}$ term is $0$,adding it to the sum of the first $18$ terms does not change the total sum. Therefore,both $S_{18}$ and $S_{19}$ equal $513$.