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. Quantos metros que ele correu no ultimo?
P.A <

Answer 1

Soma de PA = $\frac{(a1+an)n}{2}$

a11 = quantos metros correu no último dia

35200 = $\frac{(a1+a11).11}{2}$

(a1 + a11).11= 35200.2

(a1+ a11).11 = 70400

(a1 + a11) = 6400

a1 = 6400 - a11

Fórmula para achar o termo de uma PA

an = a1+ (n-1).r

A razão (r) da PA é 400

a11 = (6400 - a11) + (11-1).400

a11 = 6400 - a11 + 4000

2a11 =10400

a11 = 5200 metros

Answer 2

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

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