Average of 3-Digit AP Terms

The AP has common difference $d = 55 - 38 = 17$. General term $T_n = 38 + (n - 1)17 = 17n + 21$.

First 3-digit term ($T_n \ge 100$): $17n + 21 \ge 100 \implies 17n \ge 79 \implies n = 5$ $T_5 = 17(5) + 21 = 106$

Last 3-digit term ($T_m \le 999$): $17m + 21 \le 999 \implies 17m \le 978 \implies m = 57$ $T_{57} = 17(57) + 21 = 990$

Since terms in an AP are symmetric, the average of all terms in an AP is equal to the average of the first and last terms: $\text{Average} = \frac{\text{First term} + \text{Last term}}{2} = \frac{106 + 990}{2} = \frac{1096}{2} = 548$