Question

An arithmetic sequence has $n \ge 3$ terms, initial term $a,$ and common difference $d \gt 1.$ Carl wrote down all the terms in this sequence correctly except for one term, which was off by $1.$ The sum of the terms he wrote was $222.$ What was $a + d + n?$

$24$

$20$

$22$

$28$

$26$

Solution:

The true sum is $S = na + \frac{n(n-1)}{2}d.$ Since one term is off by $1,$ the written total satisfies $222 = S \pm 1,$ so $S = 221$ or $223.$ Also $2S = n\bigl(2a + (n-1)d\bigr).$ If $S = 223,$ which is prime, no factorization with $n \ge 3$ yields valid $a, d.$ So take $S = 221 = 13 \cdot 17.$ With $n = 13,$ we get $2a + 12d = 34,$ that is $a + 6d = 17,$ so $a = 5, d = 2,$ and $d \gt 1$ checks out. Then $a + d + n = 5 + 2 + 13 = 20.$ Thus, B is the correct answer.

2023 AMC 10B Problem 23

Solution:

Problem 23 in Other Years