Геометрична прогресия

[Гост]

[tex]b_1 = ? \\ q=?[/tex]

[tex]\begin{tabular}{|l}b_1+b_2+b_3=35 \\b_1^2+b_2^2+b_3^2=525 \end{tabular}[/tex]

[ammornil]

[tex]b_1 = ? \\ q=?[/tex]

[tex]\begin{array}{|l} b_1+b_2+b_3=35 \\ b_1^2+b_2^2+b_3^2=525 \end{array} \\
\begin{array}{|l}b_1+b_1.q+b_1.q^2=35 \\ b_1^2+b_1^2.q^2+b_1^2.q^4=525 \end{array} \hspace{12} \Rightarrow \hspace{12} \begin{array}{|l} b_1.(q^2+q+1)=35 \\ b_1^2.(q^4+q^2+1)=15.35 \end{array} \hspace{12} \Rightarrow \hspace{12} \begin{array}{|l} b_1.(q^2+q+1)=35 \\ \frac{(q^2+q+1)^2}{q^4+q^2+1}=\frac{35.35}{15.35} \end{array}[/tex]

Решаваме второто уравнение:
[tex]\frac{(q^2+q+1)^2}{q^4+q^2+1}=\frac{7}{3} \hspace{12} \Longleftrightarrow \hspace{12} 3.(q^4+q^2+1+2.q^3+2.q^2+2.q)=7.(q^4+q^2+1) \\
3.q^4+6.q^3+9.q^2+6.q+3-7.q^4-7.q^2-7=0 \\
-4.q^4+6.q^3+2.q^2+6.q-4=0 \hspace{12}|:(-2) \\
2.q^4-3.q^3-q^2-3.q+2=0 \\
2.q^4-4.q^3+q^3-2.q^2+q^2-3.q+2=0 \\
2.q^3.(q-2)+q^2.(q-2)+(q-1).(q-2)=0 \\
(q-2).(2.q^3+q^2+q-1)=0 \\
(q-2).[(q^3-1^3)+q^3+q^2+q)=0 \\
(q-2).[(q-1).(q^2+q+1)+q.(q^2+q+1)]=0 \\
(q-2).(q^2+q+1).(q-1+q)=0 \\
(q-2).(2.q-1).(q^2+q+1)=0 \\
q_1=2 \hspace{12} q_2=\frac{1}{2}[/tex]

Връщаме се към системата за да намерим [tex]b_1[/tex]
[tex]\begin{array}{|l}b_1.(q^2+q+1)=35 \\ q=2 \end{array} \hspace{12} \cup \hspace{12} \begin{array}{|l}b_1.(q^2+q+1)=35 \\ q=\frac{1}{2} \end{array} \\
\begin{array}{|l}b_1.(4+2+1)=35 \\ q=2 \end{array} \hspace{12} \cup \hspace{12} \begin{array}{|l}b_1.(\frac{1}{4}+\frac{1}{2}+1)=35 \\ q=\frac{1}{2} \end{array} \\
\begin{array}{|l}b_1=5 \\ q=2 \end{array} \hspace{12} \cup \hspace{12} \begin{array}{|l}b_1=20 \\ q=\frac{1}{2} \end{array}[/tex]

[Гост]

Molia, niakoi da mi pomogne s reshavaneto na zadachata!
Namerete a1 i d na aritmetichnata progresia, v koiato sborat na purvite 6 chlena e 12 , a sborat na sledvashtite 6 chlena e 84.

[ammornil]

[tex]\div a_1, \hspace{2}a_2, \hspace{2}a_3, \hspace{2}a_4, \hspace{2}a_5, \hspace{2}a_6, \hspace{2}a_7, \hspace{2}a_8, \hspace{2}a_9, \hspace{2}a_{10}, \hspace{2}a_{11}, \hspace{2}a_{12} \\
a_1+a_2+a_3+a_4+a_5+a_6=12 \\
a_7+a_8+a_9+a_{10}+a_{11}+a_{12}=84[/tex]

[tex]\begin{array}{|l} S_{6}=12 \\ S_{12}=96 \end{array} \hspace{12} \Rightarrow \hspace{12} \begin{array}{|l} \frac{2.a_1+5.d}{2}.6=12 \\ \frac{2.a_1+11.d}{2}.12=96 \end{array} \hspace{12} \Rightarrow \hspace{12} \begin{array}{|l} 3.(2.a_1+5.d)=12 \\ 6.(2.a_1+11.d)=96 \end{array} \hspace{12} \Rightarrow \hspace{12} \begin{array}{|l} 2.a_1+5.d=4 \\ 2.a_1+11.d=16 \end{array} \hspace{12} \Rightarrow \\
\begin{array}{|l}2.a_1+5.d=4 \\ 6.d=12 \end{array} \hspace{12} \Rightarrow \hspace{12} \begin{array}{|l}d=2 \\ 2.a_1+5.2=4 \end{array} \hspace{12} \Rightarrow \hspace{12} \begin{array}{|l}d=2 \\ 2.a_1=-6 \end{array} \hspace{12} \Rightarrow \hspace{12} \begin{array}d=2 \\ a_1=-3 \end{array}[/tex]

[Гост]

Моля за помощ за следната задача:

[tex]\begin{tabular}{|l}S_2=15 \\b_1-q = \frac{25}{3} \end{tabular}[/tex]

[tex]b_1 = ? \\ q=?[/tex]

[Гост]

|b1+b1.q=15
|3.b1-3.q=25

[Гост]

|b1+b1.q=15
|3.b1-3.q=25

Мерси много човек, напълно бях изключил вчера докато се пробвах да я решавам.