Consider an infinite geometric series with first term '$a$' and common ratio '$r$'. If the sum of infinite geometric series is 4 and the second term is $\dfrac{3}{4}$ then

The sum of an infinite geometric series is given by $S = \dfrac{a}{1-r} = 4$, where $|r| < 1$.

The second term of the series is $ar = \dfrac{3}{4}$.

From the first equation, $a = 4(1-r)$. Substituting this into the second equation:

$4(1-r)r = \dfrac{3}{4}$

$4r - 4r^2 = \dfrac{3}{4}$

$16r - 16r^2 = 3$

$16r^2 - 16r + 3 = 0$

Solving the quadratic equation using the quadratic formula $r = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$r = \dfrac{16 \pm \sqrt{256 - 192}}{32} = \dfrac{16 \pm \sqrt{64}}{32} = \dfrac{16 \pm 8}{32}$

This gives two possible values for $r$:

$r_1 = \dfrac{24}{32} = \dfrac{3}{4}$ and $r_2 = \dfrac{8}{32} = \dfrac{1}{4}$

For $r = \dfrac{3}{4}$, $a = 4(1 - \dfrac{3}{4}) = 4(\dfrac{1}{4}) = 1$.

For $r = \dfrac{1}{4}$, $a = 4(1 - \dfrac{1}{4}) = 4(\dfrac{3}{4}) = 3$.

Comparing with the given options, the pair $a=3$ and $r=\dfrac{1}{4}$ matches option (3).

Answer: $a=3 \quad r=\dfrac{1}{4}$