Question

У геометричній прогресії (bn) bn = 6,q = 0,5, Sn = 90. Знайдіть b1, n​

Answer 1

$\displaystyle\bf\\b_{n} =6\\\\q=0,5\\\\S_{n} =90\\\\b_{1} =? \ \ ; \ \ n=?\\\\\\S_{n} =\frac{b_{n} q-b_{1} }{q-1} \\\\\\90=\frac{6\cdot 0,5-b_{1} }{0,5-1} \\\\\\90=\frac{3-b_{1} }{-0,5} \\\\\\-45=3-b_{1} \\\\\\\boxed{b_{1} =48}\\\\\\b_{n} =b_{1} \cdot q^{n-1}\\\\\\6=48\cdot 0,5^{n-1}\\\\\\\Big(\frac{1}{2} \Big)^{n-1} =6:48\\\\\\\Big(\frac{1}{2} \Big)^{n-1} =\Big(\frac{1}{2} \Big)^{3} \\\\\\n-1=3 \ \ \ \Rightarrow \ \ \ \boxed{n=4}} }$