The terms of an infinitely decreasing geometric progression in which all the terms are positive, the first term is 4 , and the difference between third and fifth term is $\dfrac{32}{81}$, then which of the following is not true

Let the first term be $a = 4$ and the common ratio be $r$, where $0 < r < 1$. The terms of the geometric progression are $a, ar, ar^2, ar^3, ar^4, \dots$.

The third term is $T_3 = ar^2 = 4r^2$ and the fifth term is $T_5 = ar^4 = 4r^4$.

Given $T_3 - T_5 = \dfrac{32}{81}$, we have $4r^2 - 4r^4 = \dfrac{32}{81}$.

Dividing by 4, $r^2 - r^4 = \dfrac{8}{81}$, which implies $r^4 - r^2 + \dfrac{8}{81} = 0$.

Let $x = r^2$. Then $x^2 - x + \dfrac{8}{81} = 0$. Solving for $x$ using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$x = \dfrac{1 \pm \sqrt{1 - 4(8/81)}}{2} = \dfrac{1 \pm \sqrt{1 - 32/81}}{2} = \dfrac{1 \pm \sqrt{49/81}}{2} = \dfrac{1 \pm 7/9}{2}$.

The two possible values for $x = r^2$ are $x_1 = \dfrac{1 + 7/9}{2} = \dfrac{16/9}{2} = \dfrac{8}{9}$ and $x_2 = \dfrac{1 - 7/9}{2} = \dfrac{2/9}{2} = \dfrac{1}{9}$.

If $r^2 = \dfrac{1}{9}$, then $r = \dfrac{1}{3}$. In this case, $S_{\infty} = \dfrac{a}{1 - r} = \dfrac{4}{1 - 1/3} = \dfrac{4}{2/3} = 6$.

If $r^2 = \dfrac{8}{9}$, then $r = \sqrt{\dfrac{8}{9}} = \dfrac{2\sqrt{2}}{3}$. In this case, $S_{\infty} = \dfrac{4}{1 - 2\sqrt{2}/3} = \dfrac{12}{3 - 2\sqrt{2}}$. Rationalizing the denominator, $S_{\infty} = \dfrac{12(3 + 2\sqrt{2})}{9 - 8} = 12(3 + 2\sqrt{2}) = 36 + 24\sqrt{2}$.

Comparing with the options, $r = 1/3$ is true, $S_{\infty} = 6$ is true, $r = 2\sqrt{2}/3$ is true, and $S_{\infty} = 3 + 2\sqrt{2}$ is not true.

Answer: $S_{\infty}=3+2 \sqrt{2}$