Question

знайдіть суму перших п'яти членів геометричної прогресії у якої b2=9, b3=3

Answer 1

$\displaystyle\bf\\b_{2} =9\\\\b_{3} =3\\\\b_{3} =b_{2} \cdot q\\\\q=b_{3}:b_{2} =3:9=\frac{1}{3} \\\\\boxed{q=\frac{1}{3} }\\\\b_{2} =b_{1} \cdot q\\\\b_{1} =b_{2} :q=9:\frac{1}{3} =9\cdot 3=27\\\\\boxed{b_{1} =27}\\\\\\S_{5} =\frac{b_{1} \cdot(1-q^{5} )}{1-q} =\frac{27\cdot\Big[1-\Big(\dfrac{1}{3}\Big)^{5} \Big] }{1-\dfrac{1}{3} } =\frac{27\cdot\Big(1-\dfrac{1}{243} \Big)}{\dfrac{2}{3} } =$

$\displaystyle\bf\\=\frac{27\cdot 3\cdot\dfrac{242}{243} }{2} =\frac{\dfrac{242}{3} }{2} =\frac{242}{6} =40\frac{1}{3} \\\\\\Otvet: \ S_{5} =40\frac{1}{3}$