Question

15.6. Запишите пять первых членов геометрической прогрессии
(bn), если:
1) b, = 0,6 и q = 2;
2) b, = −1,2 и q =
1/3;
3) b, = −27 и q=-2/3;
4) b, = 3,6 и q = 1/6

Answer 1

Ответ:

Объяснение:

1) $b_{1}$=0,6

$b_{2}$=$b_{1} *q$=0,6*2=1,2

$b_{3}=b_{2}*q$=1,2*2=2,4

$b_{4}=2,4*2=4,8$

$b_{5}=4,8*2=9,6$

2)

$b_{1}=-1,2\\q=\frac{1}{3}$

$b_{2}=-1,2*\frac{1}{3}=-0,4\\b_{3}=-0,4*\frac{1}{3}=\frac{-0.4}{3}=\frac{-4}{30}\\b_{4}=\frac{-4}{30}*\frac{1}{3}==\frac{-4}{90}\\b_{5}=\frac{-4}{90}*\frac{1}{3}==\frac{-4}{270}$

3)

$b_{1}=-27\\ q=\frac{-2}{3} \\b_{2}=-27*\frac{-2}{3}=18\\b_{3}=18*\frac{-2}{3}=-12\\b_{4}=-12*\frac{-2}{3}=8\\b_{5}=8*\frac{-2}{3}=\frac{-16}{3}$

4)

$b_{1}=3,6\\ q=\frac{1}{6}\\ b_{2}=3,6* \frac{1}{6}=0,6\\ b_{3}=0,6* \frac{1}{6}=0,1\\ b_{4}=0,1* \frac{1}{6}=\frac{0,1}{6}=\frac{1}{60}\\ b_{5}=\frac{1}{60}* \frac{1}{6}=\frac{1}{360}$