Question

Um atleta corre sempre 400m a mais que o dia anterior. Ao final de 11 dias ele percorre um w total de 35200 metros. O número de metros que ele correu no último dia foi igual à:

Answer 1

Progressão Aritmética

an = a1 + (n -1).r → a11 = a1 + 10.400 = a1 + 4000 → a1 = a11 - 4000
Sn = [(a1 + an).n]/2 → 35200 = [(a11 - 4000 + a11).11]/2 → 35200 = (22.a11 - 44000)/2 → 70400 + 44000 = 22.a11 → 114400 = 22.a11 → a11 = 5200m

Answer 2

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

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