Ответ: b₁=-10 q=-2 S₄=50.
Объяснение:
[tex]\displaystyle\\\frac{b_{N+10}}{b_{N+5}} =-32\ \ \ \ \ \ b_4+b_2=10N\ \ \ \ \ \ N=10\ \ \ \ b_1 =?\ \ \ \ q=?\ \ \ \ S_4=?\\\\\\\frac{b_{10+10}}{b_{10+5}}=-32\\\\\\\frac{b_{20}}{b_{15}} =(-2)^5\\\\\\\frac{b_1q^{19}}{b_1q^{14}}=(-2)^5\\\\\\q^5=(-2)^5\ \ \ \ \Rightarrow\\\\q=-2.\\\\b_4+b_2=10*10\\\\b_1q^3+b_1q=100\\\\b_1*((-2)^3+(-2))=100\\\\b_1*(-8-2)=100\\\\-10*b_1=100\ |:(-10)\\\\b_1=-10.\\\\S_4=-10*\frac{(-2)^4-1}{-2-1} =-10*\frac{16-1}{-3} =-10*\frac{15}{-3}=-10*(-5)=50.[/tex]
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Ответ: b₁=-10 q=-2 S₄=50.
Объяснение:
[tex]\displaystyle\\\frac{b_{N+10}}{b_{N+5}} =-32\ \ \ \ \ \ b_4+b_2=10N\ \ \ \ \ \ N=10\ \ \ \ b_1 =?\ \ \ \ q=?\ \ \ \ S_4=?\\\\\\\frac{b_{10+10}}{b_{10+5}}=-32\\\\\\\frac{b_{20}}{b_{15}} =(-2)^5\\\\\\\frac{b_1q^{19}}{b_1q^{14}}=(-2)^5\\\\\\q^5=(-2)^5\ \ \ \ \Rightarrow\\\\q=-2.\\\\b_4+b_2=10*10\\\\b_1q^3+b_1q=100\\\\b_1*((-2)^3+(-2))=100\\\\b_1*(-8-2)=100\\\\-10*b_1=100\ |:(-10)\\\\b_1=-10.\\\\S_4=-10*\frac{(-2)^4-1}{-2-1} =-10*\frac{16-1}{-3} =-10*\frac{15}{-3}=-10*(-5)=50.[/tex]