\[ a_3 = a_1 + 2d \] En consecuencia tenemos: \[ a_1 + (a_1 + d) + (a_1 + 2d) = 30 \hspace{2em} \boldsymbol{(1)} \] \[ a_1 (a_1 + d) (a_1 + 2d) = 510 \hspace{3em} \boldsymbol{(2)} \]

De (1):

\[ 3 a_1 + 3d = 30 \Rightarrow a_1 + d = 10 \hspace{2em} \boldsymbol{(3)} \]

Entonces:

\[ a_1 = 10 – d \hspace{3em} \boldsymbol{(4)} \]

Reemplazando (4) en (2):

\[ \begin{aligned} &(10 – d) (10) (10 + d) = 510 \\[.5em] &\hspace{3em}\Rightarrow (10 – d) (10 + d) = 51 \\[.5em] &\hspace{3em}\Rightarrow 100 – d^2 = 51 \\[.5em] &\hspace{3em}\Rightarrow d^2 = 49 \\[.5em] &\hspace{3em}\Rightarrow d = 7 \end{aligned} \]

Reempazando \(d = 7\) en (4):

\[ a_1 = 10 – 7 = 3 \]

Luego, el mayor de los trés números es:

\[ \begin{aligned} a_3 &= a_1 + 2d \\[.5em] &= 3 + 2(7) \\[.5em] &= \boldsymbol{17} \end{aligned} \]