Question

(UFCE) Um atleta corre sempre 400 metros a mais que no dia anterior. Ao final de 11 dias ele percorre um total de 35200 metros. O número de metros que ele correu no último dia foi igual a    (A) 5100    (B) 5200    (C) 5300     (D) 5400    (E) 5500

Answer 1

Resposta

letra b) 5200

Passo a Passo:

primeira coisa vamos descobrir quanto ele correu no primeiro dia.

lembrando que a cada dia ele corre 400 metros a mais que o dia anterior.

x + (x+400)+(x+800)+(x+1200)+(x+1600)+(x+2000)+(x+2400)+(x+2800)+(x+3200)+(x+3600)+(x+4000)

11x+22000=35200

11x=35200-22000

11x = 13200

x=13200/11

x=1200

em seguida substituímos o x pelo valor encontrado e fazemos as somas.

1200 + (1200+400)+(1200+800)+(1200+1200)+(1200+1600)+(1200+2000)+(1200+2400)+(1200+2800)+(1200+3200)+(1200+3600)+(1200+4000)

1200 + 1600+2000+2400+2800+3200+3600+4000+4400+4800+5200

Answer 2

$\sf \displaystyle 35200=\left(a1+an\right)\frac{11}{2}\\\\\\\sf 35200=\left(a\cdot \:1+an\right)\frac{11}{2}\\\\\\\sf \left(a\cdot \:1+an\right)\frac{11}{2}=35200\\\\\\\sf 2\left(a\cdot \:1+an\right)\frac{11}{2}=35200\cdot \:2\\\\\\\sf 11\left(a+an\right)=70400\\\\\\\sf \frac{11\left(a+an\right)}{11}=\frac{70400}{11}\\\\\\\sf a+an=6400\\\\\\\sf a\left(1+n\right)=6400\\\\\\\sf \frac{a\left(1+n\right)}{1+n}=\frac{6400}{1+n}\\\\\\\to \boxed{\boxed{\sf a=6400}}$

$\tt an=\left(6400-an\right)+\left(11-1\right)\cdot 400\\\\\\\tt an=6400-an+\left(11-1\right)\cdot \:400\\\\\\\tt an=6400-an+4000\\\\\\\tt an=-an+10400\\\\\\\tt an+an=-an+10400+an\\\\\\\tt 2an=10400\\\\\\\tt \dfrac{2an}{2n}=\dfrac{10400}{2n}\\\\\\\to \boxed{\boxed{\sf a=5200}}$