Question

Sa se determine termenul 10 al progresiei aritmetice (an) daca:
a) a3=12 , a6=30
b)a2=48 , a24=-150
c)a25=450,r=-1/3

Dau coroana!!!!!!!!!!!

Answer 1

a)
Aplicam Formula termenului general

      { a₁+2*r=12 =>*(-1) =>-a₁-2*r=-12=>    3*r=18=>r=6
      {a₁+5*r=30 =>             a₁+5*r=30 =>
Termenul de rang 10 este
a₁₀=a₁+9*r=>a₁₀=0+9*6=>a₁₀=0+54=>a₁₀=54

b)a₂=48, a₂₄=-150

      { a₁+r=48 =>                      a₁ + r= 48=>            a₁=48-r
      {a₁+23*r=-150 =>              a₁+23*r=-150            48-r+23r=-150=>22r=-198=>r=-9  

 r=-9
a₁=48-r=48+9=>a₁=57
Termenul 10 este:
a₁₀=a₁+9*r=>a₁₀=57+9*(-9)=>a₁₀=57-81=-24

c)a₂₅=450, r=-1/3
a₂₅=a₁+24*r
450=a₁+24*(-1/3)=>a₁=450-24*(-1/3)=>a₁=458
Termenul de rang 10 este
a₁₀=a₁+9*r=>a₁₀=458+9*(-1/3)=>a₁₀=455

Answer 2

[tex]\displaystyle a).a_3=12,~a_6=30,~a_{10}=?\\a_3=12 \Rightarrow a_{3-1}+r=12\Rightarrow a_2+r=12\Rightarrow a_1+2r=12\Rightarrow\\ \Rightarrow a_1=12-2r\\a_6=30\Rightarrow a_{6-1}+r=30\Rightarrow a_5+r=30\Rightarrow a_1+5r=30\Rightarrow\\ \Rightarrow 12-2r+5r=30\Rightarrow -2r+5r=30-12\Rightarrow3r=18 \Rightarrow \\ \Rightarrow r= \frac{18}{3} \Rightarrow r=6 \\ a_1=12-2r \Rightarrow a_1=12-2 \cdot 6 \Rightarrow a_1=12-12 \Rightarrow a_1=0 [/tex]
[tex]\displaystyle a_{10}=a_{10-1}+r\Rightarrow a_{10}=a_9+r\Rightarrow a_{10}=a_1+9r \Rightarrow a_{10}=0+9 \cdot 6 \Rightarrow \\ \Rightarrow a_{10}=0+54 \Rightarrow \boxed{a_{10}=54 } \\ b).a_2=48,~ a_{24}=-150 ,~a_{10}=? \\ a_2=48 \Rightarrow a_{2-1}+r=48 \Rightarrow a_1+r=48 \Rightarrow a_1=48-r [/tex]
[tex]\displaystyle a_{24}=-150 \Rightarrow a_{24-1}+r=-150 \Rightarrow a_{23}+r=-150 \Rightarrow \\ \Rightarrow a_1+23r=-150 \Rightarrow 48-r+23r=-150 \Rightarrow \\ \Rightarrow -r+23r=-150-48 \Rightarrow 22r=-198 \Rightarrow r=- \frac{198}{22} \Rightarrow r=- 9 \\ a_1=48-r \Rightarrow a_1=48-(-9) \Rightarrow a_1=48+9 \Rightarrow a_1=57 [/tex]
[tex]a_{10}=a_{10-1}+r \Rightarrow a_{10}=a_9+r \Rightarrow a_{10}=a_1+9r \Rightarrow \\ \Rightarrow a_{10}=57+9 \cdot (-9) \Rightarrow a_{10}=57-81 \Rightarrow \boxed{a_{10}=-24} [/tex]
$\displaystyle c).a_{25}=450 ,~r=- \frac{1}{3} ,~a_{10}=? \\ a_{25}=450 \Rightarrow a_{25-1}+r=450 \Rightarrow a_{24}+r=450 \Rightarrow a_1+24r=450 \Rightarrow \\ \Rightarrow a_1+24 \cdot \left(- \frac{1}{3} \right )=450 \Rightarrow a_1- \frac{24}{3} =450 \Rightarrow a_1-8=450 \Rightarrow \\ \Rightarrow a_1=450+8 \Rightarrow a_1=458$
$\displaystyle a_{10}=a_{10-1}+r \Rightarrow a_{10}=a_9+r \Rightarrow a_{10}=a_1+9r \Rightarrow \\ \Rightarrow a_{10}= 458+9 \cdot \left(- \frac{1}{3} \right) \Rightarrow a_{10}= 458- \frac{9}{3} \Rightarrow a_{10}=458-3 \Rightarrow \\ \Rightarrow \boxed{a_{10}=455}$