Question

Sa se resolve ecuatile:
A)x+2x+3x+...+20x=3+6+9+87
B)2a+4a+6a+...+20a+586=a+3a+5a+7a+...+31a

Answer 1

a) x+2x+3x+....+20x=105
    +   +   + .......+=105
  20x  19x  18x...x
(21x X 20):2=105
210x=105

Answer 2

avem 1+2+3+...+n=$\frac{n(n+1)}{2}$
A) x+2x+3x+...+20x=3+6+9+87
$\left \{ {{S=\frac{20x(20x+1)}{2}} \atop {S=3+6+9+87}} \right. \\ \frac{20x(20x+1)}{2}=105 \\ 10x(20x+1)=105 \\ 200x+10x=105 \\ 210x=105 \\ x= \frac{105}{210} \\ x= \frac{1}{2}$

La B ma incurca ce este dupa egal...nu imi dau seama cum sa fac suma.