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

Spa=(a1+an)n/2
35200=(a1+an)11/2
35200*2=(a1+an)11
70400/11=(a1+an)
6400=(a1+an)
a1=6400-an

an=a1+(n-1)r
an=(6400-an)+(11-1).400
an=6400+4000-an
2an=10400
an=10400/2
an=5200

no ultimo dia ele percorreu 5200 metros

Answer 2

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

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