1, Ta có \(\left\{{}\begin{matrix}u_1=-1\\u_1.q=3\end{matrix}\right.\Rightarrow\dfrac{1}{q}=-\dfrac{1}{3}\Leftrightarrow q=-3\)

\(S_{10}=-1.\dfrac{1-\left(-3\right)^{10}}{1-\left(-3\right)}=14762\)

2, tương tự 

1:

\(S_8=\dfrac{u_1\cdot\left(1-q^8\right)}{1-q}=\dfrac{2048\cdot\left(1-\left(\dfrac{5}{4}\right)^8\right)}{1-\dfrac{5}{4}}\)

\(=-8192\left(1-\left(\dfrac{5}{4}\right)^8\right)\)

2:

\(u2=u1\cdot q\)

=>\(q=\dfrac{3}{-1}=-3\)

\(S_{10}=\dfrac{u1\left(1-q^{10}\right)}{1-q}=\dfrac{-1\cdot\left(1-\left(-3\right)^{10}\right)}{1-\left(-3\right)}\)

\(=\dfrac{-1}{4}\left(1-3^{10}\right)\)

1:

\(S_{10}=\dfrac{u_1\cdot\left(1-q^{10}\right)}{1-q}=\dfrac{-3\cdot\left(1-\dfrac{1}{1024}\right)}{1-\dfrac{1}{2}}\)

\(=-6\cdot\dfrac{1023}{1024}=\dfrac{-3069}{512}\)

2:

\(\left\{{}\begin{matrix}u1=6\\u2=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u1=6\\u1\cdot q=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u1=6\\q=3\end{matrix}\right.\)

\(S_{12}=\dfrac{u_1\left(1-q^{12}\right)}{1-q}=\dfrac{6\cdot\left(1-3^{12}\right)}{1-3}=-3\cdot\left(1-3^{12}\right)\)

\(=3^{13}-3\)

1: u3=-3 và u9=29

=>u1+2d=-3 và u1+8d=29

=>-6d=-32 và u1+2d=-3

=>d=16/3 và u1=-3-2d=-3-32/3=-41/3

2: \(S_{20}=\dfrac{20\cdot\left[2\cdot u1+19\cdot d\right]}{2}=10\cdot\left(-5\cdot2+19\cdot3\right)\)

=10(57-10)

=10*47=470

Câu 2:

\(\left\{{}\begin{matrix}u_1+u_5-u_3=10\\u_1+u_6=17\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u_1+u_1+4d-u_1-2d=10\\u_1+u_1+5d=17\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u_1+2d=10\\2u_1+5d=17\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2u_1+4d=20\\2u_1+5d=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2u_1+4d-2u_1-5d=20-17\\2u_1+5d=17\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-d=3\\2u_1+5d=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=-3\\2u_1=17-5d=17+5\cdot3=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u_1=16\\d=-3\end{matrix}\right.\)

Câu 1:

Để a,b,c lập thành cấp số cộng thì

\(\left[{}\begin{matrix}a+c=2b\\a+b=2c\\b+c=2a\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x+1+x^2-1=2\cdot\left(3x-2\right)\\x+1+3x-2=2\left(x^2-1\right)\\x^2-1+3x-2=2\left(x+1\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x^2+x-6x+4=0\\2x^2-2=4x-1\\x^2+3x-3-2x-2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x^2-5x+4=0\\2x^2-4x-1=0\\x^2+x-5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\left(x-1\right)\left(x-4\right)=0\\2x^2-4x-1=0\\x^2+x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\left\{1;4\right\}\\x\in\left\{\dfrac{2+\sqrt{6}}{2};\dfrac{2-\sqrt{6}}{2}\right\}\\x\in\left\{\dfrac{-1+\sqrt{21}}{2};\dfrac{-1-\sqrt{21}}{2}\right\}\end{matrix}\right.\)

a: \(\left\{{}\begin{matrix}u5-u1=15\\u4-u1=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u1\cdot q^4-u1=15\\u1\cdot q^3-u1=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u1\left(q^4-1\right)=15\\u1\left(q^3-1\right)=6\end{matrix}\right.\Leftrightarrow\dfrac{q^4-1}{q^3-1}=\dfrac{5}{2}\)

=>\(2\left(q^4-1\right)=5\left(q^3-1\right)\)

=>\(2q^4-2-5q^3+5=0\)

=>\(2q^4-5q^3+3=0\)

=>\(2q^4-2q^3-3q^3+3=0\)

=>\(2q^3\left(q-1\right)-3\left(q-1\right)\left(q^2+q+1\right)=0\)

=>\(\left(q-1\right)\left(2q^3-3q^2-3q-3\right)=0\)

=>\(\left[{}\begin{matrix}q=1\\q\simeq2,39\end{matrix}\right.\)

=>\(u1=\dfrac{6}{q^3-1}\simeq\dfrac{6}{2.39^3-1}\simeq0,47\)

b: \(\left\{{}\begin{matrix}u1-u3+u5=65\\u1+u7=325\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u1-u1\cdot q^2+u1\cdot q^4=65\\u1+u1\cdot q^6=325\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u1\cdot\left(1-q^2+q^4\right)=65\\u1\left(1+q^6\right)=325\end{matrix}\right.\)

=>\(\dfrac{1-q^2+q^4}{1+q^6}=\dfrac{65}{325}=\dfrac{1}{5}\)

=>\(\dfrac{1}{q^2+1}=\dfrac{1}{5}\)

=>\(q^2+1=5\)

=>q^2=4

=>q=2 hoặc q=-2

TH1: q=2

=>\(u1=\dfrac{325}{q^6+1}=5\)

TH2: q=-2

=>\(u1=\dfrac{325}{\left(-2\right)^6+1}=5\)

a: 

ĐKXĐ: \(q\notin\left\{0;1;-1\right\}\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}u1\cdot q^4-u1=15\\u1\cdot q^3-u1\cdot q=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{q^4-1}{q^3-q}=\dfrac{15}{6}=\dfrac{5}{2}\\u1\left(q^4-1\right)=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2q^4-2=5q^3-5q\\u1\left(q^4-1\right)=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2q^4-5q^3+5q-2=0\\u1\left(q^4-1\right)=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(q-2\right)\left(q-1\right)\left(q+1\right)\left(2q-1\right)=0\\u1\left(q^4-1\right)=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}q=2\\q=\dfrac{1}{2}\end{matrix}\right.\\u1\left(q^4-1\right)=15\end{matrix}\right.\)

TH1: q=2

=>\(u1=\dfrac{15}{2^4-1}=\dfrac{15}{15}=1\)

TH2: q=1/2

=>\(u1=\dfrac{15}{\dfrac{1}{16}-1}=15:\dfrac{-15}{16}=-16\)

b:

 \(HPT\Leftrightarrow\left\{{}\begin{matrix}u1-u1\cdot q^2+u1\cdot q^4=65\\u1+u1\cdot q^6=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{q^4-q^2+1}{q^6+1}=\dfrac{1}{5}\\u1\left(1+q^6\right)=325\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{q^2+1}=\dfrac{1}{5}\\u1\left(q^6+1\right)=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}q^2=4\\u1\left(q^6+1\right)=325\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}q\in\left\{2;-2\right\}\\u1\left(q^6+1\right)=325\end{matrix}\right.\Leftrightarrow u1=\dfrac{325}{65}=5\)

c: \(HPT\Leftrightarrow\left\{{}\begin{matrix}u1\cdot q^3+u1\cdot q^5=-540\\u1\cdot q+u1\cdot q^3=-60\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{q^5+q^3}{q^3+q}=9\\u1\left(q+q^3\right)=-60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}q^2=9\\u1\left(q+q^3\right)=-60\end{matrix}\right.\)

TH1: q=3

\(u1=-\dfrac{60}{3+3^3}=-\dfrac{60}{30}=-2\)

TH2: q=-3

=>\(u1=-\dfrac{60}{-3-27}=\dfrac{60}{30}=2\)