Question

n(x+1)+(x+2)+(x+3)+...+(x+20)=250

(x+2)+(x+4)+(x+6)+...+(x+20)=220

 

Answer 1

a)(x+1)+(x+2)+(x+3)+...+(x+20)=250

x+1+x+2+x+3+...+x+20 = 250

20x+(1+2+3+...+20) = 250

20x + 210 = 250

20x          = 250 - 210

20x          = 40

x             = 40 : 20

x             = 2

b)(x+2)+(x+4)+(x+6)+...+(x+20)=220

x+2+x+4+x+6+...+x+20=220

10x + (2+4+6+...+20)=220

10x + 110 = 220

10x       = 220 - 110

10x       = 110

x          = 110 : 10

x          = 11

Answer 2

(x + 1) + (x + 2) + ... + (x + 20) = 250

=> 20x + (1 + 2 + 3 + ... + 20) = 250

=> 20x + \(\frac{20.21}{2}\)= 250

=> 20x + 210 = 250

=> 20x = 40

=> x = 2

(x + 2) + (x + 4) + ... + (x + 20) = 220

=> 10x + (2 + 4 + ... + 20) = 220

=> 10x + \(\frac{\left[\left(20-2\right):2+1\right].\left(20+2\right)}{2}\)= 220

=> 10x + 110 = 220

=> 10x = 110

=> x = 11