Найдите s5 если для геометрической прогрессии s2=4 s3=13

$S_n= \frac{b_1(q^n-1)}{q-1} \\\ \left \{ {{S_2= \frac{b_1(q^2-1)}{q-1} =4} \atop {S_3= \frac{b_1(q^3-1)}{q-1} =13}} \right. \\\ \left \{ {{\frac{b_1(q-1)(q+1)}{q-1} =4} \atop {\frac{b_1(q-1)(q^2+q+1)}{q-1} =13}} \right. \\\ \left \{ {{b_1(q+1) =4} \atop {b_1(q^2+q+1) =13}} \right. \\\ b_1= \frac{4}{q+1} = \frac{13}{q^2+q+1} \\\ 4q^2+4q+4=13q+13 \\\ 4q^2-9q-9=0 \\\ D=9^2+4\cdot4\cdot9=225 \\\ q= \frac{9+15}{8} =3; \ b_1= \frac{4}{3+1} =1 \\\ q= \frac{9-15}{8}=-0.75; \ b_1= \frac{4}{-0.75+1} =16$
$S_5= \frac{b_1(q^5-1)}{q-1} \\\ S_5= \frac{1\cdot(3^5-1)}{3-1} =121 \\\ S_5= \frac{16\cdot((- \frac{3}{4} )^5-1)}{- \frac{3}{4} -1} =\frac{16\cdot(- \frac{243}{1024}-1)}{- \frac{7}{4} } =\frac{16\cdot(- \frac{1267}{1024})}{- \frac{7}{4} } =\frac{ \frac{1267}{64}}{ \frac{7}{4} } =\frac{1267}{112}=\frac{181}{16}$
Ответ: 121 или 181/16