Question

um atleta corre sempre 400 metros a mais que no dia anterior ao final de 11 dias ele percorre um total de 352000 metros determine o numero de metros q ele correu no ultimo dia

Answer 1

a11 = a1 + (n + r)
a11 = a1 + (11 - 1) . 400
a11 = a1 + 10 . 400
a11 = a1 + 4000
a1 = a11 - 4000

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Formula da soma:

Sn = ( a1 + an ) . n /  2
35200 = ( a11 - 4000 + a11) . 11 / 2 
35200 = (a22 - 4000) . 11 
35200 . 2 = (a22 - 4000) . 11
70400 = 22a11 - 44000
70400 + 44000 = 22a11
114400 = 22a11
22a11 = 114400
a11 = 114400 / 22
a11 = 5200 metros

Answer 2

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

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