a) Ta có:

Vậy số hạng đầu u₁ = 8, công sai d = -3

b) Ta có:

(1) ⇔ 2u₁ + 20d = 60 ⇔ u₁ = 30 – 10d thế vào (2)

(2) ⇔[(30 – 10D) + 3d]² + [(30 – 10d) + 11d]² = 1170

⇔ (30 – 7d)² + (30 + d)² = 1170

⇔900 – 420d + 49d² + 900 + 60d + d² = 1170

⇔ 50d² – 360d + 630 = 0

Vậy

hay

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

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

a) \(\left\{{}\begin{matrix}u_5=96\\u_7=384\end{matrix}\right.\)

\(u^2_6=u_5.u_7=96.384=36864\)

\(\Leftrightarrow u_6=192\)

\(q=\dfrac{u_7}{u_6}=\dfrac{384}{192}=2\)

\(u_5=u_1.q^4\)

\(\Leftrightarrow u_1=\dfrac{u_5}{q^4}=\dfrac{96}{2^4}=6\)

b) \(\left\{{}\begin{matrix}u_4-u_2=25\\u_3-u_1=50\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_1.q^3-u_1.q=25\\u_1.q^2-u_1=50\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_1.q\left(q^2-1\right)=25\left(1\right)\\u_1.\left(q^2-1\right)=50\left(2\right)\end{matrix}\right.\)

\(\left(1\right):\left(2\right)\Leftrightarrow q=\dfrac{25}{50}=\dfrac{1}{2}\)

\(\left(2\right)\Leftrightarrow u_1=\dfrac{50}{q^2-1}=\dfrac{50}{\dfrac{1}{4}-1}=-\dfrac{200}{3}\)

a)

\(\begin{array}{l}\left\{ \begin{array}{l}5{u_1} + 10{u_5} = 0\\{S_4} = 14\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}5{u_1} + 10\left( {{u_1} + 4{\rm{d}}} \right) = 0\\\frac{{4\left( {2{u_1} + 3{\rm{d}}} \right)}}{2} = 14\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}5{u_1} + 10{u_1} + 40{\rm{d}} = 0\\2\left( {2{u_1} + 3{\rm{d}}} \right) = 14\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}15{u_1} + 40{\rm{d}} = 0\\2{u_1} + 3{\rm{d}} = 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 8\\d =  - 3\end{array} \right.\end{array}\)

Vậy cấp số cộng \(\left( {{u_n}} \right)\) có số hạng đầu \({u_1} = 8\) và công sai \(d =  - 3\).

b)

\(\begin{array}{l}\left\{ \begin{array}{l}{u_7} + {u_{15}} = 60\\u_4^2 + u_{12}^2 = 1170\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left( {{u_1} + 6{\rm{d}}} \right) + \left( {{u_1} + 14{\rm{d}}} \right) = 60\\{\left( {{u_1} + 3{\rm{d}}} \right)^2} + {\left( {{u_1} + 11{\rm{d}}} \right)^2} = 1170\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1} + 6{\rm{d}} + {u_1} + 14{\rm{d}} = 60\\{\left( {{u_1} + 3{\rm{d}}} \right)^2} + {\left( {{u_1} + 11{\rm{d}}} \right)^2} = 1170\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}2{u_1} + 20{\rm{d}} = 60\\{\left( {{u_1} + 3{\rm{d}}} \right)^2} + {\left( {{u_1} + 11{\rm{d}}} \right)^2} = 1170\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1} + 10{\rm{d}} = 30\left( 1 \right)\\{\left( {{u_1} + 3{\rm{d}}} \right)^2} + {\left( {{u_1} + 11{\rm{d}}} \right)^2} = 1170\left( 2 \right)\end{array} \right.\end{array}\)

\(\left( 1 \right) \Leftrightarrow {u_1} = 30 - 10{\rm{d}}\) thế vào (2) ta được:

\(\begin{array}{l}{\left( {30 - 10{\rm{d}} + 3{\rm{d}}} \right)^2} + {\left( {30 - 10{\rm{d}} + 11{\rm{d}}} \right)^2} = 1170 \Leftrightarrow {\left( {30 - 7{\rm{d}}} \right)^2} + {\left( {30 + {\rm{d}}} \right)^2} = 1170\\ \Leftrightarrow 900 - 420{\rm{d}} + 49{{\rm{d}}^2} + 900 + 60{\rm{d}} + {d^2} = 1170 \Leftrightarrow 50{{\rm{d}}^2} - 360{\rm{d}} + 630 = 0\\ \Leftrightarrow 5{{\rm{d}}^2} - 36{\rm{d}} + 63 = 0 \Leftrightarrow \left[ \begin{array}{l}d = 3\\d = \frac{{21}}{5}\end{array} \right.\end{array}\)

Với \(d = 3 \Leftrightarrow {u_1} = 30 - 10.3 = 0\).

Với \(d = \frac{{21}}{5} \Leftrightarrow {u_1} = 30 - 10.\frac{{21}}{5} =  - 12\).

Vậy có hai cấp số cộng \(\left( {{u_n}} \right)\) thoả mãn:

‒ Cấp số cộng có số hạng đầu \({u_1} = 0\) và công sai \(d = 3\).

‒ Cấp số cộng có số hạng đầu \({u_1} =  - 12\) và công sai \(d = \frac{{21}}{5}\).

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

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

1: u2=4 và u4=10

=>u1+d=4 và u1+3d=10

=>2d=6 và u1+d=4

=>d=3 và u1=1

\(S_{10}=\dfrac{10\cdot\left(2\cdot1+9\cdot3\right)}{2}=5\cdot\left(2+27\right)=145\)

2: 

u3=6 và u5=16

=>u1+2d=6 và u1+4d=16

=>2d=10  và u1+2d=6

=>d=5 và u1=6-2*5=-4

\(S_{12}=\dfrac{12\cdot\left(2\cdot\left(-4\right)+11\cdot5\right)}{2}=6\cdot\left(-8+55\right)=6\cdot47=282\)

a) \(\left\{{}\begin{matrix}u_2-u_3+u_5=10\\u_4+u_6=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+d-u_1-2d+u_1+4d=10\\u_1+3d+u_1+5d=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+3d=10\\2u_1+8d=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1=1\\d=3\end{matrix}\right.\)

b)\(\left\{{}\begin{matrix}u_2-u_6+u_4=-7\\u_8-2u_7=2u_4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+d-u_1-5d+u_1+3d=-7\\u_1+7d-2\left(u_1+6d\right)=2\left(u_1+3d\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1-d=-7\\-3u_1-11d=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1=\dfrac{-11}{2}\\d=\dfrac{3}{2}\end{matrix}\right.\)

c)\(\left\{{}\begin{matrix}u_7-u_3=8\\u_2.u_7=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+6d-u_1-2d=8\\\left(u_1+d\right)\left(u_1+6d\right)=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4d=8\\\left(u_1+d\right)\left(u_1+6d\right)=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=2\\\left(u_1+2\right)\left(u_1+12\right)=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=2\\u_1^2+14u_1+24=75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=2\\\left[{}\begin{matrix}u_1=3\\u_1=-17\end{matrix}\right.\end{matrix}\right.\)

Gọi số hạng đầu và công sai của cấp số cộng lần lượt là: 1u_1 và d.
Ta có:
154111\left\{{}\begin{matrix}u_1+2u_5=0\\S_4=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+2.\left(u_1+4d\right)=0\\\dfrac{\left[2u_1+3d\right].4}{2}=14\end{matrix}\right.11\Leftrightarrow\left\{{}\begin{matrix}3u_1+8d=0\\2u_1+3d=7\end{matrix}\right.1\Leftrightarrow\left\{{}\begin{matrix}u_1=8\\d=-3\end{matrix}\right..

b) Gọi số hạng đầu và công sai của cấp số cộng làn lượt là \(u_1\) d. Ta có:
\(\left\{{}\begin{matrix}u_1+3d=10\\u_1+6d=19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u_1=1\\d=3\end{matrix}\right.\).
c) Gọi số hạng đầu và công sai của cấp số cộng lần lượt là \(u_1\) và d. Ta có:
\(\left\{{}\begin{matrix}u_1+u_1+4d-u_1-2d=10\\u_1+u_1+5d=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u_1+2d=10\\2u_1+5d=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u_1=36\\d=-13\end{matrix}\right.\).
d) Gọi số hạng đầu và công sai của cấp số cộng lần lượt là \(u_1\) và d. Ta có:
\(\left\{{}\begin{matrix}u_1+6d-\left(u_1+2d\right)=8\\\left(u_1+d\right)\left(u_1+6d\right)=75\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}4d=8\\\left(u_1+d\right)\left(u_1+6d\right)=75\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}d=2\\\left(u_1+2\right)\left(u_1+12\right)=75\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}d=2\\u^2_1+14u_1-51=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}d=\\\left[{}\begin{matrix}u_1=3\\u_1=-17\end{matrix}\right.\end{matrix}\right.\)
Vậy có hai cấp số cộng thỏa mãn là: \(\left\{{}\begin{matrix}d=2\\u_1=3\end{matrix}\right.\) và \(\left\{{}\begin{matrix}d=2\\u_1=-17\end{matrix}\right.\).