Solution da2e3374-85e4-4b34-968b-1b95981d2e87

Points: 3 pts

Lemma 1: $n\bigl(2a+(n-1)d\bigr)=442$ or $446$.

Let $S$ be the true sum. Using the arithmetic sequence sum formula,
\[
S=\sum_{k=0}^{n-1}(a+kd)
 =na+\frac{n(n-1)}{2}d
 =\frac{n}{2}\bigl(2a+(n-1)d\bigr).
\]
Since exactly one term was off by $1$, the true sum is $221$ or $223$.
Thus we obtain the Diophantine equation
\[
n\bigl(2a+(n-1)d\bigr)=442
\quad\text{or}\quad 446.
\]

Lemma 2: $n=13$.

Since $a\geq 1$ and $d\geq 2$, we have
\[
n^2=\frac{n}{2}\bigl(2+2(n-1)\bigr)\leq S\leq 223.
\]
Therefore $3\leq n\leq 14$. Factoring the two possible right-hand sides gives
\[
442=2\cdot 13\cdot 17,
\qquad
446=2\cdot 223.
\]
The integer $n$ must divide one of these numbers. The only divisor in the
range $3\leq n\leq 14$ is $13$, which divides $442$. Hence $n=13$.

Lemma 3: $d=2$, $a=5$.

Substituting $n=13$ into the Diophantine equation gives
\[
13(2a+12d)=442
\quad\Longrightarrow\quad
 a+6d=17.
\]
Because $a\geq 1$,
\[
17=6d+a\geq 6d+1
\quad\Longrightarrow\quad
 d\leq \frac{8}{3}\approx 2.67.
\]
Together with $d\geq 2$, we see $d=2$ and $a=5$. Therefore,
\[
a+d+n=5+2+13=20.
\]

Final answer:
20