S3=28 q=1/2 S6=?.Алгебра

Решите задачу:

$S_3=28\; ,\; \; q=\frac{1}{2}\; ,\; \; S_6=?\\\\S_3=\frac{b_1(1-q^3)}{1-q}\; ,\; \; \; S_6=\frac{b_1(1-q^6)}{1-q}\\\\\frac{b_1}{1-q}=\frac{S_3}{1-q^3}=\frac{S_6}{1-q^6}\; \; \to \; \; \frac{1-q^6}{1-q^3}=\frac{S_6}{S_3}\; ,\; \; \frac{1-( \frac{1}{2})^6}{1-( \frac{1}{2})^3}=\frac{S_6}{28}\\\\\frac{1-\frac{1}{64}}{1-\frac{1}{8}}=\frac{S_6}{28}\; ,\; \; \frac{63/64}{7/8}=\frac{S_6}{28}\; ,\; \; \frac{63\cdot 8}{64\cdot 7}=\frac{S_6}{28} \; ,\; \; \frac{9}{8}=\frac{S_6}{28}\; ,\\\\S_6=\frac{9\cdot 28}{8}=31,5$