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:

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

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

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

a/ \(u_6=u_1+5d=8\Rightarrow u_1=8-5d\)

\(u_2=u_1+d;u_4=u_1+3d\)

\(\Rightarrow\left\{{}\begin{matrix}u_2=8-5d+d=8-4d\\u_4=8-5d+3d=8-2d\end{matrix}\right.\)

\(\Rightarrow\left(8-4d\right)^2+\left(8-2d\right)^2=16\Rightarrow...\)

b/ Câu này làm theo ý hiểu thôi, ko chắc đâu

\(Xet-S_n:\)

\(u_1=u_1\)

\(u_2=u_1+d\)

\(u_3=u_1+2d\)

......

\(u_n=u_1+\left(n-1\right)d\)

\(\Rightarrow S_n=u_1+u_2+...+u_n=u_1+u_1+d+...+u_1+\left(n-1\right)d=n.u_1+d+2d+....+\left(n-1\right)d\)

\(=n.u_1+\left(1+2+...+\left(n-1\right)\right)d=n.u_1+\dfrac{d\left(n-1\right).n}{2}=\dfrac{n\left[2u_1+\left(n-1\right)d\right]}{2}\)

Tương tụ với S(2n)

\(S_{2n}=u_1+u_2+...+u_{2n}=u_1+u_1+d+....+u_1+\left(2n-1\right)d\)

\(=2n.u_1+d+2d+...+\left(2n-1\right)d=2n.u_1+\left(1+2+...+\left(2n-1\right)\right)d=2n.u_1+d.n\left(2n-2\right)=2n\left(u_1+\left(n-1\right).d\right)\)

\(4S_n=S_{2n}\Leftrightarrow4.\dfrac{n\left[2u_1+\left(n-1\right)d\right]}{2}=2n\left(u_1+\left(n-1\right).d\right)\)

\(\Leftrightarrow2n\left[2u_1+\left(n-1\right)d\right]=2n\left[u_1+\left(n-1\right)d\right]\)\(\Leftrightarrow2u_1=u_1\Rightarrow u_1=0\)

\(u_5=u_1+4d=18\Rightarrow d=\dfrac{18}{4}=4,5\)

Ok check lại số má hộ tui nhó

camon bn

a: u1-2u4+u6=12 và u2+u5=8

=>u1-2u1-6d+u1+5d=12 và u1+d+u1+4d=8

=>d=12 và 2u1+5d=8

=>d=12 và 2u1=8-5d=8-60=-52

=>u1=-26 và d=12

b: u5-u2=3 và u3*u8=24

=>u1+4d-u1-d=3 và (u1+2d)(u1+7d)=24

=>d=1 và (u1+2)(u1+7)=24

=>d=1 và u1^2+9u1-10=0

=>d=1 và (u1=-10 hoặc u1=1)

a: u4=4 và u6=8

=>u1+3d=4 và u1+5d=8

=>-2d=-4 và u1+3d=4

=>d=2 và u1=4-3d=-2

b: u1-u3+u5=10 và u1+u6=17

=>u1-u1-2d+u1+4d=10 và u1+u1+5d=17

=>u1+2d=10 và 2u1+5d=17

=>u1=16 và d=-3

c: u1+u2=5 và u3*u5=91

=>u1+u1+d=5 và (u1+2d)(u1+4d)=91

=>2u1+d=5 và (u1+2d)(u1+4d)=91

=>d=5-2u1 và (u1+10-4u1)(u1+20-8u1)=91

=>d=5-2u1 và (-3u1+10)(-7u1+20)=91

(-3u1+10)(-7u1+20)=91

=>21u1^2-60u1-70u1+200=91

=>21u1^2-130u1+109=0

=>u1=1 hoặc u1=109/21

Khi u1=1 thì d=5-2u1=5-2=3

Khi u1=109/21 thì d=5-2u1=5-218/21=-113/21

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

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

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

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

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

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

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:

\(u2=u1\cdot q\)

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

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

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

a.

\(\left\{{}\begin{matrix}u_1+\left(u_1+4d\right)-\left(u_1+2d\right)=10\\\left(u_1+d\right)+\left(u_1+4d\right)=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u_1+2d=10\\2u_1+5d=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=36\\d=-13\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}u_1+d+u_1+3d=5\\u_1^2+\left(u_1+4d\right)^2=25\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4d=5-2u_1\\u_1^2+\left(u_1+4d\right)^2=25\end{matrix}\right.\)

\(\Rightarrow u_1^2+\left(u_1+5-2u_1\right)^2=25\)

\(\Rightarrow u_1^2+u_1^2-10u_1+25=25\)

\(\Rightarrow\left[{}\begin{matrix}u_1=0\Rightarrow d=\dfrac{5}{4}\\u_1=5\Rightarrow d=-\dfrac{5}{4}\end{matrix}\right.\)

Lỗi công thức toán rồi bạn.

\(\left\{{}\begin{matrix}S_{15}=585\\u_1^3+\left(u_2\right)^3=302094\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}15\cdot\dfrac{2\cdot u_1+14d}{2}=585\\u_1^3+\left(u_1+d\right)^3=302094\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u_1+7d=39\\u_1^3+\left(u_1+d\right)^3=302094\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u_1=39-7d\\\left(39-7d\right)^3+\left(39-7d+d\right)^3=302094\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u_1=39-7d\\59319-31941d^2+5733d-343d^3+59319-18252d^2+2808d-216d^3=302094\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_1=39-7d\\118638-50193d+8541d^2-559d^3=302094\end{matrix}\right.\)

=>d=-2,46(loại)

Vậy: Không có bộ số số hạng đầu và công sai nào thỏa mãn đề bài