Răspuns :

Madalin01vaicarid

Uite partea teoretica a rezolvarii tale:

-formula termenului general: [tex]a_{n}=a_{1}+(n-1)*r[/tex]

-suma primilor n termeni: [tex]S_{n}=\frac{(a_{1}+a_{n})*n}{2}[/tex]

Sₙ=2n²-3n

insa Sₙ=(a₁+aₙ)*n/2

=> (a₁+aₙ)*n/2=2n²-3n => (a₁+aₙ)*n=2(2n²-3n) => (a₁+aₙ)*n=4n²-6n

=> (a₁+aₙ)*n=n*(4n-6) => a₁+aₙ=4n-6 <=> a₁+a₁+(n-1)*r=4n-6 => 2a₁+n*r-r=4n-6

=> 4n=r*n si 2a₁-r=-6

4n=r*n => r=4

2a₁-r=-6 <=> 2a₁-4=-6 => 2a₁=-6+4 => 2a₁=-2 => a₁=-1

aₙ=-1+4*(n-1)

Targoviste44id

[tex]\it a_1=S_1=2-3=-1\\ \\ a_1+a_2=S_2=2\cdot4-6=2 \Rightarrow -1+a_2=2 \Rightarrow a_2=3\\ \\ r=a_2-a_1=3-(-1)=3+1=4\\ \\ a_n=a_1+(n-1)r=-1+(n-1)\cdot4=-1+4n-4=4n-5[/tex]