Consider an infinite geometric series with first term $a$ and common ratio $r$. If its sum is 4 and the second term is $3/4$, then

For an infinite geometric series, the sum is $$\frac{a}{1-r}=4.$$ Also, the second term is $$ar=\frac{3}{4}.$$ From the sum equation, $$a=4(1-r).$$ Substitute into the second condition: $$4r(1-r)=\frac{3}{4}.$$ So, $$16r-16r^2=3.$$ Rearranging, $$16r^2-16r+3=0.$$ Factoring gives $$ (4r-1)(4r-3)=0. $$ Hence $$r=\frac{1}{4}$$ or $$r=\frac{3}{4}.$$ If $$r=\frac{1}{4},$$ then $$a=4\left(1-\frac{1}{4}\right)=3.$$ If $$r=\frac{3}{4},$$ then $$a=4\left(1-\frac{3}{4}\right)=1,$$ which is not among the options. Comparing with the given choices, the matching pair is option $D$.