The sequence \(6;6;9;15;\ldots\) has a constant second difference of 3 thus it is a quadratic sequence.

The first differences are \(0,3,6,\dots\) respectively.

To get the general term of a quadratic sequence we need to use the formula

\[T_n = an^2+bn+c\]

where \(a;b;c\) are obtained from the following simultaneous equations:

\(2a=3\)

\(\therefore a=\dfrac{3}{2}\)

\(5a + b = 3\)

\(5(\frac{3}{2})+b=3\)

\(b=\dfrac{-9}{2}\)

\(4a+2b+c=6\)

\( 4(\frac{3}{2})+2(\frac{-9}{2})+c=6\)

\(\therefore c=9\)

Substituting \(a,b,c\) in \(T_n=an^2+bn+c\) gives:

\((\frac{3}{2})25+(\frac{-9}{2})5+9\)

\(=\dfrac{75-45+18}{2} = 24\)