We are given an arithmetic sequence of positive integers that contains the numbers 28, 52, and 82 as terms. The sum of all terms in the sequence is 1769. We need to find the smallest and largest terms in the sequence.

Step 1: General form of an arithmetic sequence

The general form of an arithmetic sequence is:

$a_n = a_1 + (n-1) d$

where:

• $a_n$ is the $n$-th term,
• $a_1$ is the first term,
• $d$ is the common difference.

Step 2: Equation setup using known terms

Let’s assume the position of 28, 52, and 82 in the sequence are $a_i$, $a_j$, and $a_k$ respectively. Therefore:

$a_i = a_1 + (i-1) d = 28$ $a_j = a_1 + (j-1) d = 52$ $a_k = a_1 + (k-1) d = 82$

From these, we can subtract the equations to find $d$.

Subtract the first equation from the second:

$(a_1 + (j-1) d) - (a_1 + (i-1) d) = 52 - 28$ $(j-i)d = 24 \quad \text{so} \quad d = \frac{24}{j-i}$

Subtract the second equation from the third:

$(a_1 + (k-1) d) - (a_1 + (j-1) d) = 82 - 52$ $(k-j) d = 30 \quad \text{so} \quad d = \frac{30}{k-j}$

Now we have two equations for $d$:

1. $d = \frac{24}{j-i}$
2. $d = \frac{30}{k-j}$

Step 3: Solving for positions

Let’s try to find the smallest integer values for $j - i$ and $k - j$ that make the common difference $d$ consistent in both equations.

Assume $j - i = 2$. Then:

$d = \frac{24}{2} = 12$

Substitute $d = 12$ into the second equation:

$30 = (k-j) \cdot 12 \quad \Rightarrow \quad k - j = \frac{30}{12} = 2.5$

Since $k - j$ must be an integer, this assumption leads to a contradiction. Therefore, try different values for $j - i$.

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