Question

1)Se considera o progresie aritmetica $a_{n}$ n ≥1 in care a3=5si a6=11, Sa se calculeze a9.
2)Sa se calculeze suma 1+2+$2^{2}$+$2^{3}$+......+$2^{7}$.
3)Se considera progresia aritmetica $a_{n}$ n$a_{n \geq }$1 in care $a_{1}$= 1 si $a_{5}$=13.Sa se calculeze $a_{2009}$.
Vaaa rooog!!!!

Answer 1

1)

[tex]a_6=a_3+3r \Longrightarrow 3r=a_6-a_3=11-5=6\\\;\\ a_9=a_6+3r=11+6=17[/tex]

2)

$1+2+2^2+2^3+ ... +2^7 = \dfrac{2^8-1}{2-1} =255$

3)

[tex]a_5=a_1+4r \Rightarrow r=\dfrac{a_5-a_1}{4} =\dfrac{13-1}{4} =3 \\\;\\ a_{2009}=a_1+2008r =1+2008\cdot3[/tex]

Answer 2

$\displaystyle 1).a_3=5,~a_6=11,~a_9=? \\ a_3=5 \Rightarrow a_{3-1}+r=5 \Rightarrow a_2+r=5 \Rightarrow a_1+2r=5 \Rightarrow a_1=5-2r \\ a_6=11 \Rightarrow a_{6-1}+r=11 \Rightarrow a_5+r=11 \Rightarrow a_1+5r=11 \Rightarrow \\ \Rightarrow 5-2r+5r=11 \Rightarrow -2r+5r=11-5 \Rightarrow 3r=6 \Rightarrow r= \frac{6}{3} \Rightarrow r=2 \\ a_1=5-2r \Rightarrow a_1=5-2 \cdot 2 \Rightarrow a_1=5-4 \Rightarrow a_1=1$
$\displaystyle a_9=a_{9-1}+r \Rightarrow a_9=a_8+r \Rightarrow a_9=a_1+8r \Rightarrow a_9=1+8 \cdot 2 \Rightarrow \\ \Rightarrow a_9=1+16 \Rightarrow \boxed{a_9=17}$
$\displaystyle 2).1+2+2^2+2^3+...+2^7 \\ b_1=1,~b_2=2,~b_3=2^2,~b_4=2^3 \\ q= \frac{b_2}{b_1} \Rightarrow q= \frac{2}{1} \Rightarrow q=2 \\ S_{8}=1 \cdot \frac{2^8-1}{2-1} =2^8-1=256-1=\boxed{255}$
$\displaystyle 3).a_1=1,~a_5=13,~a_{2009}=? \\ a_5=13 \Rightarrow a_{5-1}+r=13 \Rightarrow a_4+r=13 \Rightarrow a_1+4r=13 \Rightarrow \\ \Rightarrow 1+4r=13 \Rightarrow 4r=13-1 \Rightarrow 4r=12 \Rightarrow r= \frac{12}{4} \Rightarrow r=3 \\ a_{2009}=a_{2009-1}+r \Rightarrow a_{2009}=a_{2008}+r \Rightarrow a_{2009}=a_1+2008r \Rightarrow \\ \Rightarrow a_{2009}=1+2008 \cdot 3 \Rightarrow a_{2009}=1+6024 \Rightarrow \boxed{a_{2009}=6025}$