Question

Um atleta corre sempre 400metros a mais que no dia anterior .Ao final de 11 dias ele percorre um total de 35200metros. O numero de metros que ele correu no último dia foi igual a?

Answer 1

x2 = x1 + 400

x3 = x2 + 400 = x1 + 400 + 400 = x1 + 2*400

x4 = x3 + 400 = x1 + 400 + 400 + 400

...

xN = x1 + (N-1)*400                          

para uma PA, a soma de todos os termos é igual á

Spa = $\frac{(x1 + xN)*N}{2}$

aplicando a equação

$35200 = \frac{(x1 + <strong>x11</strong>)*11}{2}$

substituindo a equação para xN na equação acima:

35200 = $\frac{(x1 + <strong>x1 + (11-1)*400</strong>)<strong>*11</strong>}{2}$

resolvendo essa equação

x1 = 1200

dessa forma

xN tem que ser igual a 5200             (1200 + 10*400)

Answer 2

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

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