Question

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:

Answer 1

Montando uma PA

[ a1, a1 + 400, ..., a11 ] 

an = a1 + [ n - 1 ] . r

a11 = a1 + [ 11 - 1 ] . 400

a11 = a1 + 10 . 400 ou an = a1 + 10 . 4 000

a11 = a1 + 4 000

Soma = 35 200

Sn = [ a1 + an ] . n / 2

35 200 = [ a1 + a1 + 4 000 ] . 11 / 2

2 . 35 200 = [ 2a1 + 4 000 ] . 11

70 400 = 22a1 + 44 000

70 400 - 44 000 = 22a1

22a1 = 26 400

a1 = 26 400 / 22

a1 = 1 200

a11 = a1 + 4 000

a11 = 1 200 + 4 000

a11 = 5 200 metros

ou

an = 5 200 metros 

Answer 2

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

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