Question

3- (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:

B

Explicação passo-a-passo:

Pelo enunciado, podemos ver que se trata de um problema de P.A., em que:

razão = 400

n de termos = 11

Soma = 35200

S = (a1 + an) .n/2

35200 = (a1 + an).11/2

70400/11 = a1 + an

6400 = a1 + an

a1 = 6400 - an

an = a1 + (n-1).r

an = 6400 -an + 400.10

an + an = 6400 + 4000

2an = 10400

an = 5200

Answer 2

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

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