Question

Sa se rezolve ecuatiile:
a) (3x-1)+(3x-4)+ (3x-7)+..+(3x-58)=790

Answer 1

  
[tex]\displaystyle (3x-1)+(3x-4)+ (3x-7)+..+(3x-58)=790 \\ \text{Calculam numarul de termeni ai sumei: } \\ \\ n = \frac{58-1}{3}+1 = \frac{57}{3}+1 =19+1 = 20 ~de~termeni \\ \\ (3x-1)+(3x-4)+ (3x-7)+..+(3x-58)=790 \\ 3x+3x+3x+ \hdots de ~20~de~ori~ \hdots+3x -1-4-7- \hdots - 58 = 790\\ 20\cdot 3x - (1+4+7+ \hdots 58) = 790 \\ \\ 20\cdot 3x - \frac{20(58+1)}{2} =790 \\ 20\cdot 3x - 10 \cdot 59 =790 \\ 60x - 590=790\\ 60x=790 +590 \\ 60x= 1380 [/tex]


[tex]\displaystyle x= \frac{1380}{60} = \boxed{23}[/tex]