Решите пожалуйста..... Created by timurchik1497. algebra-ru

1.3.20с₄=с₁q³=√5c₇=c₁q⁶=-25 1.3.21. 1.3.22 1.3.23 не удовлетворяет условию 1.3.24q=-3

Решите пожалуйста...

timurchik1497 @timurchik1497

July 2022 1 10 Report

Решите пожалуйста....

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Yena

1.3.20

с₄=с₁q³=√5

c₇=c₁q⁶=-25

$\frac{c_7}{c_4}=\frac{c_1q^6}{c_1q^3}=-\frac{25}{\sqrt{5}}$

$q^3=-(\sqrt{5})^3$

$q=-\sqrt{5}$

1.3.21.

$P_n=(b_1*b_n)^\frac{n}{2}$

$b_1=\frac{1}{16}$

$q=\frac{1}{4}:\frac{1}{16}=4$

$b_7=\frac{1}{16}*4^6=256$

$P_7=(\frac{1}{16}*256)^\frac{7}{2}=16^\frac{7}{2}=16384$

1.3.22

$b_1=-\frac{1}{9}$

$q=\frac{1}{3}:(-\frac{1}{9})=-3$

$b_6=-\frac{1}{9}*(-3)^5=27$

$P_6=(-(\frac{1}{9})*27)^\frac{6}{2}=(-3)^3=-27$

1.3.23

$P_4=(2*b_4)^\frac{4}{2}=(2*2*q^3)^2=4^2*q^6=16*q^6=1024$

$q^6=64$

$q=2$ $q=-2$ не удовлетворяет условию

1.3.24

$P_5=((-\frac{1}{27})*b_5)^\frac{5}{2}=((-\frac{1}{27})*(-\frac{1}{27})*q^4)^\frac{5}{2}=(\frac{1}{729}*q^4)^\frac{5}{2}=-\frac{1}{243}$

$(\frac{1}{729})^\frac{5}{2}*(q^4)^\frac{5}{2}=-\frac{1}{243}$

$\frac{1}{27^5}*q^{10}=-\frac{1}{243}$

$q^{10}=-\frac{1}{243}*27^5$

$q^{10}=-3^{10}$

q=-3

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