Question

Знайдіть суму чотирьох перших членів геометричної про гресії (bn) зі знаменником q, якщо:
1) b4=100; q=4;
$2)b1 = 2 \sqrt{2} .b7 = 16 \sqrt{2} .q > 0$
3) b2=12, b5=324​

Answer 1

$\displaystyle\bf\\1)\\\\b_{4} =100 \ \ \ ; \ \ \ q=4 \ \ ; \ S_{4} =?\\\\\\b_{4} =b_{1}\cdot q^{3}\\\\b_{1}=b_{4}:q^{3} =100:4^{3} =100:64=1,5625\\\\\\S_{4} =\frac{b_{4}\cdot q-b_{1} }{q-1} =\frac{100\cdot 4-1,5625}{4-1}=\frac{398,4375}{3} =132,8125$

$\displaystyle\bf\\2)\\\\b_{1} =2\sqrt{2} \ \ \ , \ \ \ b_{7} =16\sqrt{2} \\\\b_{7}=b_{1}\cdot q^{6}\\\\q^{6} =b_{7} : b_{1} =16\sqrt{2} :2\sqrt{2}=8\\\\\boxed{q=\sqrt[6]{8} =\sqrt{2} }\\\\\\S_{4}=\frac{b_{1}\cdot(q^{4} -1) }{q-1} =\frac{2\sqrt{2}\cdot(4-1) }{\sqrt{2} -1} =\frac{6\sqrt{2} }{\sqrt{2}-1 }$

$\displaystyle\bf\\3)\\\\b_{2}=12 \ \ \ ; \ \ \ b_{5}=324\\\\\\:\left \{ {{b_{1} \cdot q^{4}=324 } \atop {b_{1}\cdot q=12 }} \right. \\---------\\q^{3}= 27\\\\\boxed{q=3}\\\\b_{1}=12:3=4\\\\\\S_{4} =\frac{b_{1}\cdot(q^{3}-1) }{q-1}=\frac{4\cdot(27-1)}{3-1} =\frac{4\cdot 26}{2}=52$