If S is the sum to infinity of a decreasing geometric progression with common ratio $x$ such that $|x| <1 ; x \neq 0$. The ratio of fourth term to the second term is $\dfrac{1}{16}$ and the ratio of the third term to the square of the second term is $\dfrac{1}{9}$. Then the value of S is

Let the first term of the geometric progression be $a$ and the common ratio be $x$. The terms are $a, ax, ax^2, ax^3, \dots$.

Given the ratio of the fourth term to the second term is $\dfrac{1}{16}$, we have $\dfrac{ax^3}{ax} = x^2 = \dfrac{1}{16}$. Since the progression is decreasing and $|x| < 1$, we take $x = \dfrac{1}{4}$ (as $x = -\dfrac{1}{4}$ would lead to an alternating series, but the problem implies a standard decreasing GP).

Given the ratio of the third term to the square of the second term is $\dfrac{1}{9}$, we have $\dfrac{ax^2}{(ax)^2} = \dfrac{ax^2}{a^2x^2} = \dfrac{1}{a} = \dfrac{1}{9}$. Thus, $a = 9$.

The sum to infinity $S$ of a geometric progression is given by $S = \dfrac{a}{1 - x}$.

Substituting the values $a = 9$ and $x = \dfrac{1}{4}$, we get $S = \dfrac{9}{1 - 1/4} = \dfrac{9}{3/4} = 9 \times \dfrac{4}{3} = 12$.

Answer: 12