Verified answer

[tex] > resolucao \\ \\ \geqslant progressao \: \: aritmetica \\ \\ a20 = a1 + 19r \\ 106 = 11 + 19r \\ 106 - 11 = 19r \\ 95 = 19r \\ r = \frac{95}{19} \\ r = 5 \\ \\ \\ \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant \leqslant \geqslant [/tex]

[tex]\Large\boxed{\begin{array}{l}\underline{\rm Progress\tilde{a}o~Aritm\acute{e}tica}\\\huge\boxed{\boxed{\boxed{\boxed{\sf a_n=a_1+(n-1)\cdot r}}}}\\\sf a_n\longrightarrow o~ termo ~n~na~ sequ\hat{a}ncia\\\sf a_1\longrightarrow o~ primeiro ~ termo ~na ~sequ\hat{e}ncia\\\sf r\longrightarrow a~ diferenc_{\!\!,}a ~ comum ~ entre ~ termos\\\sf Dados:\\\Longrightarrow\sf a_1=11\\\Longrightarrow\sf a_{20}=106\\\Longrightarrow\sf n=20\\\Longrightarrow\sf r=\,?\end{array}}[/tex]

[tex]\Large\boxed{\begin{array}{l}\underline{\bf Soluc_{\!\!,}\tilde{a}o:}\\\iff\sf 106=11+(20-1)\cdot r\\\\\iff\sf 106=11+19\cdot r\\\\\iff\sf 95=19r\\\\\iff\sf r=\dfrac{95}{19}\\\\\red{\iff \sf r=5}~\blue\checkmark\end{array}}[/tex]