Question

Care poate sa ma ajute.​

Answer 1

Răspuns:

Explicație pas cu pas:

2) S₅ = ? a₁ = 1 ; a₂ = 3 ; progresie aritmetica =>

r = a₂-a₁ = 3-1 = 2  => r = 2

a₃ = a₂+r = 3+2 = 5 ; a₄ = a₃+r = 5+2 = 7 ; a₅ = a₄+r = 7+2 = 9

S₅ = a₁+a₂+a₃+a₄+a₅ = 1+3+5+7+9 = 25 => S₅ = 25

3) b₄ = ? progresie geometrica ; q = 1/3 ; b₁ = 27 =>

b₄ = b₁·q³ = 27·(1/3)³ = 27·1/27 = 27/27 = 1 => b₄ = 1

4) S = 1 + 1/3 + 1/3² + 1/3³ + 1/3⁴ = (3⁴+3³+3²+3¹+3⁰)/3⁴ =

= (81+27+9+3+1)/81 = 121/81 = (11/9)²

5) x-3 ; 4 ; x+3 termeni consecutivi ai  unei progresii aritmetice

<=> 4 = (x-3+x+3)/2 = > 8 = 2x => x = 8/2 => x = 4

Termenii devin : 1 ; 4 ; 7 cu ratia r = 3

6) S = 1+3+5+7+9+....+19 = (1+19)·[(19-1):2+1] :2 = 20·10:2 = 10² = 100

[(19-1):2+1] = numarul de termeni ai sumei = 18:2+1 = 9+1 = 10

S = 100

7) a₃ = 5 ; a₆ = 11  ; a₉ = ? ; progresie aritmetica

a₃ = a₁+2r = 5 => a₁ = 5-2r

a₆ = a₁+5r = 11 => a₁ = 11-5r =>

5-2r = 11-5r => 5r-2r = 11-5 => 3r = 6 => r = 2

a₁ = 5-2·2 = 5-4 = 1

a₉ = a₁+8r = a₆+3r

a₉ = 1+8·2 = 1+16 = 17

sau a₉ = 11+3·2 = 11+6 = 17 =>

a₉ = 17

Answer 2

$\it 2.\\ \\ r=a_2-a_1=3-1=2;\ \ a_5=a_1+4r=1+4\cdot2=9\\ \\ S_5=\dfrac{(a_1+a_5)\cdot5}{2}=\dfrac{(1+9)\cdot5}{2}=\dfrac{10\cdot5}{2}=25$

$\it 3.\\ \\ b_1=27,\ \ q=\dfrac{1}{3}=3^{-1}\\ \\ \\ b_4=b_1\cdot q^3=27\cdot(3^{-1})^3=3^3\cdot3^{-3}=3^{3+(-3)}=3^0=1$

$\it 4.\\ \\ 1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}=\dfrac{1-\Big(\dfrac{1}{3}\Big)^5}{1-\dfrac{1}{3}}=\dfrac{1-\dfrac{1}{243}}{\dfrac{2}{3}}=\dfrac{242}{243}\cdot\dfrac{3}{2}=\dfrac{121}{81}$

$\it 5.\\ \\ \dfrac{x-3+x+3}{2}=4 \Rightarrow \dfrac{2x}{2}=4 \Rightarrow x=4$

$\it 6.\\ \\ 1+3+5+\ ...\ +19=10^2=100$