The average of all 3-digit terms in the arithmetic progression $38, 55, 72, \dots$, is

First term: $a_1 = 38$

Second term: $a_2 = 55$

Common difference: $d = 55 - 38 = 17$

General term formula: $a_n = 38 + (n-1) \times 17 = 21 + 17n$

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For 3-digit numbers, we need: $100 \leq a_n \leq 999$

Substituting our formula: $100 \leq 21 + 17n \leq 999$

Finding the lower bound:

$100 \leq 21 + 17n$

$79 \leq 17n$

$n \geq \frac{79}{17} = 4.65...$

Since $n$ must be a whole number, $n \geq 5$.

Finding the upper bound:

$21 + 17n \leq 999$

$17n \leq 978$

$n \leq \frac{978}{17} = 57.53...$

Since $n$ must be a whole number, $n \leq 57$.

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First 3-digit term: $a_5 = 21 + 17(5) = 21 + 85 = 106$

Last 3-digit term: $a_{57} = 21 + 17(57) = 21 + 969 = 990$

Number of 3-digit terms: $57 - 5 + 1 = 53$ terms

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In any arithmetic progression, the average equals the mean of the first and last terms.

This is because the terms are evenly spaced, so the middle value represents the average.

Average = $\frac{\text{First term} + \text{Last term}}{2} = \frac{106 + 990}{2} = \frac{1096}{2} = 548$

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The average of all 3-digit terms in the arithmetic progression is 548.

This method works for any arithmetic progression - we only need the first and last terms to find the average, regardless of how many terms there are.