Question

Fie $(a_{n})_{n\ \textgreater \ 1}$ progresia aritmetica in care se cunosc $a_{4}$ = 10 si $a_{7}$ = 19.
a) Sa se determine $a_{1}$ si ratia;
b) Sa se calculeze $S_{27}$;
c) Sa se rezolve ecuatia $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + x = 0$.

Answer 1

a₄  = a₁ + 3·r  =10 
a₇  = a₁  + 6·r  =19 
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scadere  a₁ + 6·r - a₁ -3·r = 19 - 10     ;             3·r = 9    ; r= 9 :3 = 3 
r = 3 
a₁ = 10 - 3r = 10 - 3· 3 = 10 - 9 =1 
a₁ = 1 
a₂₇ =a₁ + 26·r = 1 + 26·3 = 1 + 78 = 79 
S₂₇ = ( a₁ + a₂₇ ) · 27 /2  = ( 1 + 79 ) · 27 /2 = 80 · 27  /2 = 27 · 80 :2 =
         = 27 · 40 = 1080 
c.    a₅ = a₁  +4r =  1 + 4· 3 = 13 
a₁ + a₂ + a₃ +a₄ +a₅ = S₅ = ( a₁ + a₅ ) · 5 / 2  = ( 1 + 13 ) · 5 / 2 
  = 14  · 5 / 2  = 5 · 7 = 35 
35 + x = 0 
x = -35 

Answer 2

$\displaystyle a_4=10;~a_7=19 \\ a).a_4=10 \Rightarrow a_{4-1}+r=10 \Rightarrow a_3+r=10\Rightarrow a_1+3r=10 \Rightarrow \\ \Rightarrow a_1=10-3r \\ a_7=19 \Rightarrow a_{7-1}+r=19 \Rightarrow a_6+r=19 \Rightarrow a_1+6r=19 \Rightarrow \\ \Rightarrow 10-3r+6r=19 \Rightarrow 10+3r=19 \Rightarrow 3r=19-10 \Rightarrow 3r=9 \Rightarrow \\ \Rightarrow r= \frac{9}{3} \Rightarrow r=3 \\ a_1= 10-3r \Rightarrow a_1=10-3 \cdot 3 \Rightarrow a_1=10-9 \Rightarrow a_1=1$
$\displaystyle b).S_{27}= \frac{2+26 \cdot 3}{2} \cdot 27 \Rightarrow S_{27}= \frac{2+78}{2} \cdot 27\Rightarrow S_{27}= \frac{80}{2} \cdot 27 \Rightarrow \\ \\ \Rightarrow S_{27}=40 \cdot 27 \Rightarrow S_{27}=1080$
$\displaystyle c).a_1+a_2+a_3+a_4+a_5+x=0 \\ a_1=1 \\ a_2=a_{2-1}+r \Rightarrow a_2=a_1+r \Rightarrow a_2=1+3 \Rightarrow a_2=4 \\ a_3=a_{3-1}+r \Rightarrow a_3=a_2+1 \Rightarrow a_3=4+3 \Rightarrow a_3=7 \\ a_4=a_{4-1}+r\Rightarrow a_4=a_3+r\Rightarrow a_4=7+3 \Rightarrow a_4=10 \\ a_5=a_{5-1}+r \Rightarrow a_5=a_4+r \Rightarrow a_5=10+3 \Rightarrow a_5=13 \\ 1+4+7+10+13+x=0 \\ 35+x=0 \\ x=0-35 \\ x=-35$