Question

Daca stie careva sa rezolve asta imi trebuie la teza va rog.. Puneti o poza in com va rog

Answer 1

avem exercitii cu progrsii artmetice si progresii geometrice
1)x?
x,2x,1-x⇒progresie aritmetica
pa⇒b=a+b|2
2x=x+1-x|2⇒2x=1|2⇒2(2x)=1⇒4x=1⇒x=1|4
1|4  2 ×1|4  1-1|4
1|4,1|2 3|4
r=a2-a1
1|2-1|4⇒2-1|3⇒r=1|4
sunt cele 3 numere intr o progresie aritmetica

2)a2=2
a4=-2
a10=?
S10=?

a4=a1+3r⇒-2=2+3r⇒-2-2=3r⇒ r=
-4|3
a10=2+9(-4|3)⇒2-36|3⇒6-36|3⇒-30|3⇒a10=-10

S10=(a1+a10)x10|2⇒(2-10)x10|2⇒-80|2⇒S10=-40

ex nr3
b1=3
q=-1
b8=?
S8?

bn=b1xqⁿ⁻¹
b8=b1x(-1)⁸⁻¹⇒3x(-1)⇒-3
b8=-3

S8=3x(-1)⁸-1|(-1)-1⇒3x1-1|-2⇒3-1|-2⇒2|-2⇒S8=-1
succes si bafta la teza !
sper sa intelegi
poti intimpina aceste tipuri de cerinte si la bac!

Answer 2

1) La o progresie aritmetică, diferența a doi termeni consecutivi este constantă.

$\it 2x-x=1-x-2x \Leftrightarrow x = 1-3x|_{+3x} \Leftrightarrow 4x=1 \Leftrightarrow x=\dfrac{1}{4}$

2)

$\it a_4=a_2+2r \Leftrightarrow -2 = 2 +2r|_{-2} \Leftrightarrow -4=2r|_{:2} \Leftrightarrow -2=r \Leftrightarrow r=-2$

[tex]\it a_{10}= a_2+8r=2+8\cdot(-2) = 2-16=-14. \\\;\\ a_1=a_2-r=2-(-2) =2+2=4[/tex]

$\it S_{10} = \dfrac{(a_1+a_{10})\cdot10}{2} = (4-14)\cdot5 = -10\cdot5 = -50.$

3)

$\it b_8=b_1\cdot q^7 =3\cdot(-1)^7=3\cdot(-1)=-3$

$\it S_8= b_1\cdot \dfrac{q^8-1}{q-1} = 3\cdot \dfrac{(-1)^8-1}{-1-1}=3\cdot\dfrac{1-1}{-2} =3\cdot0=0$

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