Resposta :
[tex]i=2\%\ a.m.=\frac{1}{50}\ a.m.\\ M=3C[/tex]
[tex]M=C(1+it)\ \therefore\ \boxed{t=\dfrac{1}{i}\bigg(\dfrac{M}{C}-1\bigg)}[/tex]
[tex]t=\dfrac{1}{\frac{1}{50}}\bigg(\dfrac{3C}{C}-1\bigg)=\dfrac{50}{1}\bigg(2\bigg)\ \therefore\ \boxed{t=100\ meses}[/tex]