Question

арифметическая прогрессия
решить все 3 номера

Answer 1

$28)\ \ \ b_1+b_2+b_3=28\ \ ,\ \ \ b_4+b_5+b_6=3,5\ \ \ ,\ \ \ b_8=?\\\\b_1+b_1q+b_1q^2=28\ \ \ \to \ \ \ b_1\cdot (1+q+q^2)=28\\\\b_1q^3+b_1q^4+b_1q^5=3,5\ \ \ \to \ \ \ b_1q^3\cdot (1+q^2+q^3)=3,5\\\\\dfrac{b_1\cdot (1+q+q^2)}{b_1q^3\cdot (1+q+q^2)}=\dfrac{28}{3,5}\\\\\\\dfrac{1}{q^3}=8\ \ ,\ \ \ q^3=\dfrac{1}{8}\ \ \ \to \ \ \ q=\dfrac{1}{2}\\\\\\b_1\cdot (1+q+q^2)=b_1\cdot \Big(1+\dfrac{1}{2}+\dfrac{1}{4}\Big)=b_1\cdot \dfrac{7}{4}=28\ \ ,\ \ \ b_1=\dfrac{28\cdot 4}{7}=16\ \,$

$b_8=b_1q^7=16\cdot \dfrac{1}{2^7}=\dfrac{2^4}{2^7}=\dfrac{1}{2^3}=\dfrac{1}{8}$

$29)\ \ \ b_1+b_5=51\ \ ,\ \ \ b_2+b_6=102\ \ ,\ \ \ S_{n}=3069\ \ ,\ \ \ n=?\\\\\\\left\{\begin{array}{l}b_1+b_1q^4=51\\b_1q+b_1q^5=102\end{array}\right\ \ \left\{\begin{array}{l}b_1\cdot (1+q^4)=51\\b_1q\cdot (1+q^4)=102\end{array}\right\ \ \left\{\begin{array}{l}b_1\cdot (1+q^4)=51\\q\cdot 51=102\end{array}\right\\\\\\\left\{\begin{array}{l}b_1\cdot (1+2^4)=51\\q=2\end{array}\right\ \ \left\{\begin{array}{ccc}b_1=3\\q=2\end{array}\right$

$S_{n}=\dfrac{b_1\, (1-q^{n})}{1-q}=\dfrac{3\cdot (1-2^{n})}{1-2}=3069\ \ ,\ \ \ 3\cdot (1-2^{n})=-3069\ \ ,\\\\\\(1-2^{n})=-1023\ \ \ ,\ \ \ 2^{n}=1+1023\ \ ,\ \ 2^{n}=1024\ \ ,\ \ 2^{n}=2^{10}\ \ ,\\\\n=10$

$30)\ \ \ C_{30}^2=\dfrac{30\cdot 29}{2!}=\dfrac{30\cdot 29}{2}=15\cdot 29=435$