Let $$A = \left\{a_{1}, a_{2}, ...a_{i}, ... \right\}$$ be an arithmetic progression, and let $$B = \left\{b_{1}, b_{2}, ...b_{i}, ... \right\}$$ be a geometric progression.
The commom difference for A is 2.
The commom 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 $$a_{1}$$?

According to the question,

A = $${a_1,\ a_1+2,\ a_1+4,\ a_1+6,...}$$

B = $$\left\{0.8,\ 0.8\left(0.2\right),\ 0.8\left(0.2\right)^2,\ 0.8\left(0.2\right)^3...\right\}$$

$$a_1b_1=0.8a_1$$

$$a_2b_2=\left(a_1+2\right)0.8\times\ 0.2$$ 

$$=\left(a_1\right)0.8\times\ 0.2+2\left(0.8\right)\left(0.2\right)$$

$$a_3b_3=\left(a_1\right)0.8\times \left(0.2\right)^2+4\left(0.8\right)\left(0.2\right)^2$$

$$a_4b_4=\left(a_1\right)0.8\times\ \left(0.2\right)^3+6\left(0.8\right)\left(0.2\right)^3$$

Let S be the infinite sum of  $$a_{i}b_{i}$$

$$S =$$ $$0.8a_1$$ + $$\left(a_1\right)0.8\times\ 0.2+2\left(0.8\right)\left(0.2\right)$$ + $$\left(a_1\right)0.8\times \left(0.2\right)^2+4\left(0.8\right)\left(0.2\right)^2$$ + $$\left(a_1\right)0.8\times\ \left(0.2\right)^3+6\left(0.8\right)\left(0.2\right)^3$$+...$$\infty\ $$

Multiplying this equation by $$0.2$$, we get

$$0.2S =$$ $$0.8a_1( 0.2)$$ + $$\left(a_1\right)0.8\times\ (0.2)^2$$ + $$2\left(0.8\right)\left(0.2^2\right)$$ + $$\left(a_1\right)0.8\times \left(0.2\right)^3$$ + $$ 4\left(0.8\right)\left(0.2\right)^3$$ +...$$\infty\ $$

Subtracting, we get

$$0.8S = 0.8a_1+ 2(0.8)(0.2)+ 2(0.8)(0.2)^2 + 2(0.8)(0.2)^3 +... $$ $$\ \infty\ $$

This is an infinite GP with common ratio 0.2
$$0.8S = 0.8a_1$$ + $$\ \dfrac{\ 2\left(0.8\right)\left(0.2\right)}{1-0.2}$$

Given, $$S = 1$$

So the equation would become

$$0.8 = 0.8a_1 + 0.4$$

$$a_1 = $$ $$\dfrac{1}{2}$$