8,. Найдите q....

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В геометрической прогрессии S4=10целых 5/8 , S5=42целых 5/8 , b1=1/8,. Найдите q.

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NNNLLL54

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Ответ: q=4 .

$S_4=10\dfrac{5}{8}=\dfrac{85}{8}\ \ ,\ \ S_5=42\dfrac{5}{8}=\dfrac{341}{8}\ \ ,\ \ \ b_1=\dfrac{1}{8}\\\\\\S_4=\dfrac{b_1\, (q^4-1)}{q-1}\ \ ,\ \ \ \ \ \dfrac{\dfrac{1}{8}\, (q^4-1)}{q-1}=\dfrac{85}{8}\ \ ,\ \ \ \ \dfrac{q^4-1}{q-1}=85\ \ ,\\\\\\\dfrac{(q-1)(q+1)(q^2+1)}{q-1}=85\ \ ,\ \ \ (q+1)(q^2+1)=85\ \to \\\\\\\underline{q^3+q^2+q+1=85\ }\\\\\\S_5=\dfrac{b_1\, (q^5-1)}{q-1}\ \ ,\ \ \ \dfrac{\dfrac{1}{8}\, (q^5-1)}{q-1}=\dfrac{341}{8}\ \ ,\ \ \ \ \dfrac{q^5-1}{q-1}=341\ \ ,$

$\dfrac{(q-1)(q^4+q^3+q^2+q+1)}{q-1}=341\ \ ,\ \ \ q^4+\underbrace{q^3+q^2+q+1}_{85}=341\\\\\\q^4+85=341\ \ ,\ \ q^4=256\ \ \ \Rightarrow \ \ \ q^2-256=0\ \ ,\\\\(q^2-16)(q^2+16)=0\ \ ,\ \ \ (q-4)(q+4)(q^2+16)=0\ \ \Rightarrow \ \ q=\pm 4\ \ \\\\q=4\ \ ,\ \ S_4=\dfrac{\dfrac{1}{8}\, (4^4-1)}{4-1}=\dfrac{85}{8}\ \ \Rightarrow \ \ \ \boxed {q=4}\\\\q=-4\ \ ,\ \ S_4=\dfrac{\dfrac{1}{8}\, ((-4)^4-1)}{-4-1}\ne \dfrac{85}{8}$

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