Ответ: S₅=10.
Объяснение:
[tex]\displaystyle\\b_{N+8}:b_{N+3}=-32\ \ \ \ \ \ b_1=\frac{N}{11} \ \ \ \ \ \ N=10\ \ \ \ \ \ S_5=?\\\\b_1=\frac{10}{11} .\\\\\\\frac{b_{10+8}}{b_{10+3}} =-32\\\\\\\frac{b_{18}}{b_{13}}=(-2)^5 \\\\\\\frac{b_1q^{17}}{b_1q^{12}}=(-2)^5\\\\\\q^5=(-2)^5\ \ \ \ \ \Rightarrow\\\\q=-2.\\\\S_5=b_1*\frac{q^5-1}{q-1} =\frac{10}{11} *\frac{(-2)^5-1}{-2-1} =\frac{10*(-32-1)}{11*(-3)}=\frac{10*(-33)}{-33}=10.[/tex]
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Ответ: S₅=10.
Объяснение:
[tex]\displaystyle\\b_{N+8}:b_{N+3}=-32\ \ \ \ \ \ b_1=\frac{N}{11} \ \ \ \ \ \ N=10\ \ \ \ \ \ S_5=?\\\\b_1=\frac{10}{11} .\\\\\\\frac{b_{10+8}}{b_{10+3}} =-32\\\\\\\frac{b_{18}}{b_{13}}=(-2)^5 \\\\\\\frac{b_1q^{17}}{b_1q^{12}}=(-2)^5\\\\\\q^5=(-2)^5\ \ \ \ \ \Rightarrow\\\\q=-2.\\\\S_5=b_1*\frac{q^5-1}{q-1} =\frac{10}{11} *\frac{(-2)^5-1}{-2-1} =\frac{10*(-32-1)}{11*(-3)}=\frac{10*(-33)}{-33}=10.[/tex]