Question

Sa se rezolve ecuațiile:
a)1+4+7+....+x=117
b)1+8+15+...+x=3075
c)(3x-1)+(3x-4)+(3x-7)+....+(3x-58)=790

Answer 1

a)1+4+7+....+x=117
2+((n-1) ×3)/ ×n=117

(-1+3n) ×n=234

3n²-n-234=0

Δ=532

N=9

x=a1+8r=25

x=25

 

b) 1+8+15+....+x=3075
1+(1+7)+(1+7×2)+....+x=3075
x=1+7n
1+(1+7)+(1+7×2)+....+(1+7n)=3075
1×(n+1)+7(1+2+....+n)=3075
n+1+7×n×(n+1)/2=3075
2n+2+7n²+7n=6150
7n²+9n-6148=0
7n²-203n+212n-6148=0
7n(n-29)+212(n-29)=0
(n-29)(7n+212)=0
n=29 
x=1+7×29=204

x = 204

 

c) (3x - 1) + (3x - 4) + (3x - 7) +...+ (3x - 58)=790
(3x + 3x + 3x +.......de 20 de ori.....+ 3x ) - (1 + 4 + 7 + .... + 58) = 790
20∙ 3x - (20 × 59 / 2 ) = 790
60x - (10 ×59 )=790
60x - 590 = 790
60x = 790 + 590
60x = 1380
x = 1380 : 60 = 23
x = 23