Let $A = \{a_1, a_2, ..., a_i, ...\}$ be an arithmetic progression, and let $B = \{b_1, b_2, ..., b_i, ...\}$ be a geometric progression.

The common difference for A is 2.

The common ratio for B is 0.2, and $b_1 = 0.8$

The infinite sum of the products $a_i b_i$ is 1, where $i = 1, 2, 3, ...$

What is $a_i$?

Given information:

Progression	Type	Details
A	Arithmetic	Common difference = 2
B	Geometric	Common ratio = 0.2, first term $b_1 = 0.8$

The general term for arithmetic progression A is: $a_i = a_1 + (i-1) \times 2$

The general term for geometric progression B is: $b_i = 0.8 \times (0.2)^{i-1}$

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The infinite sum condition states: $\sum_{i=1}^{\infty} a_i b_i = 1$

Substituting the expressions:

$\sum_{i=1}^{\infty} [a_1 + 2(i-1)] \times 0.8 \times (0.2)^{i-1} = 1$

Simplifying $a_1 + 2(i-1) = a_1 - 2 + 2i$:

$\sum_{i=1}^{\infty} (a_1 - 2) \times 0.8 \times (0.2)^{i-1} + \sum_{i=1}^{\infty} 2i \times 0.8 \times (0.2)^{i-1} = 1$

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For the first sum (geometric series):

$(a_1 - 2) \times 0.8 \times \frac{1}{1-0.2} = (a_1 - 2) \times 0.8 \times \frac{1}{0.8} = a_1 - 2$

For the second sum, using the formula $\sum_{i=1}^{\infty} i \times x^{i-1} = \frac{1}{(1-x)^2}$:

$2 \times 0.8 \times \frac{1}{(1-0.2)^2} = 1.6 \times \frac{1}{(0.8)^2} = 1.6 \times \frac{1}{0.64} = 2.5$

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Combining both parts:

$(a_1 - 2) + 2.5 = 1$

$a_1 + 0.5 = 1$

$a_1 = 0.5 = \frac{1}{2}$

Since the question asks for $a_1$ (the first term of the arithmetic progression), the answer is $\frac{1}{2}$.