Question

Să se rezolve ecuația 3x+(3x+2)+(3x+4)+…..+(3x+100)=2652
(progresie aritmetica)

Answer 1

3x+(3x+2)+(3x+4)+...+(3x+100)=2652

(3x+0)+(3x+2)+(3x+4)+...(3x+100)=2652

3x+3x+...+3x+0+2+4+6+...+100=2652

51(3x)+2(0+1+2+3+...+50)=2652

51(3x)+2(50×51)/2=2652

51(3x)+2550=2652

51(3x)=102

3x=2

x=2/3

Answer 2

$3x + 3x + 2 +... + 3x + 100 = 2652 \\ 51 \times 3x + 2 + 4 + 6 + ... + 100 = 2652 \\ 153x + 2(1 + 2 + 3 + ... + 50) = 2652 \\ 153x + 2 \times \tfrac{50 \times 51}{2} = 2652 \\ 153x + 2550 = 2652 \\ 153x = 102 \\ \Leftrightarrow x=\tfrac{2}{3}$