In a decreasing geometric progression, the $$2^{\text{nd}}$$ term is 6. The sum of all infinite terms of the progression is one-eighth of the sum to infinity of the squares of the terms. The sum of the $$1^{\text{st}}$$ and the $$4^{\text{th}}$$ terms is $$\frac{p}{q}$$ where $$p, q$$ are relatively prime to each other. Then the value of $$\left[\frac{p}{q}\right]$$, where $$[x]$$ represents the greatest integer not exceeding $$x$$ is

With first term $$A$$ and ratio $$r$$, the sum of infinite terms of a geometric progression is

$$\dfrac{A}{1-r}$$

From the statement in the question we can write

$$\dfrac{A}{1-r} = \dfrac{1}{8} \cdot \dfrac{A^2}{1-r^2}$$ 

$$A = 8(1 + r)$$

We are also given that second term is $$6$$

$$\implies Ar = 6$$ 

Substituting in the above equation we get 

$$4r^2 + 4r - 3 = 0 \implies r = \dfrac{1}{2}  \text{ or}  r = \dfrac{-3}{2} $$

For the sum of infinite terms to be a finite value, we need $$|r|<1$$

$$r = \dfrac{1}{2}$$ and $$A = 12$$. 

The first and fourth terms add to 

$$A+ Ar^3 = 12\left(1+\dfrac{1}{8}\right) = \dfrac{27}{2}$$, so $$\left[\dfrac{p}{q}\right] = \left[13.5\right] = 13$$.