[tex]\bf \begin{array}{ccrllll}
term&year&\%\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
1&2000&31\\
2&2001&31+2.4\\
3&2002&(31+2.4)+2.4\\
4&2003&(31+2.4+2.4)+2.4\\
...&...&...\\
\boxed{?}&\boxed{?}&67
\end{array}
\\ \\
the\ n^{th}\textit{ term of an arithmetic sequence, will be}
[/tex][tex]\bf a_n=a_1+(n-1)d\qquad
\begin{cases}
a_1=\textit{first term}\to 31\\
d=\textit{common difference}\to 2.4\\
n=the\ n^{th}\ term
\end{cases}
\\ \quad \\
thus
\\ \quad \\
a_n=a_1+(n-1)d\implies 67=31+(n-1)2.4[/tex]
solve for "n"
the value of the "n"th term is 67, we know that much,
so, "n" will give you how many years it took to be 67
