Question

Запишіть формулу вn члена геометричної прогресії, якщо в3 +в5 =180 і в1 +в3 =20.

Answer 1

Ответ:

$\left\{\begin{array}{l}b_3+b_5=180\\b_1+b_3=20\end{array}\right\ \ \left\{\begin{array}{l}b_1q^2+b_1q^4=180\\b_1+b_1q^2=20\end{array}\right\ \ \left\{\begin{array}{l}b_1q^2\cdot (1+q^2)=180\\b_1\cdot (1+q^2)=20\end{array}\right$

$\left\{\begin{array}{l}\ 1+q^2=\dfrac{180}{b_1q^2}\\1+q^2=\dfrac{20}{b_1}\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{180}{b_1q^2}=\dfrac{20}{b_1} \\1+q^2=\dfrac{20}{b_1}\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{180\, b_1}{20\, b_1q^2}=1\\1+q^2=\dfrac{20}{b_1}\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{9}{q^2}=1\\1+q^2=\dfrac{20}{b_1}\end{array}\right$

$\left\{\begin{array}{l}9=q^2\\1+9=\dfrac{20}{b_1}\end{array}\right\ \ \left\{\begin{array}{l}q=\pm 3\\\\b_1=2\end{array}\right\ \ \ \ \Rightarrow \\\\\\a)\ \ q=-3\ ,\ b_1=2\\\\b_{n}=b_1\cdot q^{n-1}\ \ ,\ \ \ b_{n}=2\cdot (-3)^{n-1}=(-1)^{n-1}\cdot 2\cdot 3^{n-1}\\\\b)\ \ q=3\ \ ,\ \ b_1=2\\\\b_{n}=2\cdot 3^{n-1}\\\\Otvet:\ \ b_{n}=(-1)^{n-1}\cdot 2\cdot 3^{n-1}\ \ ,\ \ b_{n}=2\cdot 3^{n-1}\ .$