Question

(Unicamp) se (a1, a2,..., a13) é uma progressão aritmética (P.A) cuja soma dos termos é 78, então a7 é igual a
a) 6
b) 7
c) 8
d) 9
e) 10

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Answer 1

$\large\boxed{\begin{array}{l}\rm (a_1,a_2,a_3,\dotsc a_{13})\\\rm S_{13}=78\\\rm\dfrac{13\cdot(a_1+a_{13})}{2}=78\\\rm a_1+a_{13}=\dfrac{2\cdot\diagdown\!\!\!\!\!\!78}{\diagdown\!\!\!\!\!\!13}\\\\\rm a_1+a_{13}=2\cdot6\\\rm a_1+a_{13}=12\\\rm a_1+a_1+12r=12\\\rm 2a_1+12r=12\div2\\\rm a_1+6r=6\\\rm a_7=a_1+6r\\\rm portanto\\\rm a_7=6\\\huge\boxed{\boxed{\boxed{\boxed{\rm\dagger\red{\maltese}~\blue{alternativa~a}}}}}\end{array}}$

Answer 2

S13 = 78

Sn = (a1+an).n/2

S13 = (a1+a13).13/2

78.2/13= (a1+a13)

6.2 = a1+a13

12 = a1+a1+12r

12 = 2a1+12r

12 = 2.(a1+6r)

12/2 = a1+6r

6 = a1+6r

6 = a7

a7 = 6

R.: a)