Question

Sabendo que a soma dos cinco primeiros termos de uma P.A. é 60 e a soma dos 8 primeiros é 90. Calcule o valor do sétimo termo dessa progressão.

Answer 1

S5  = 60
S8 = 90


5( a1 + a5)/2 = 60
5(a1 + a5) =   120
5(a1 + a1 + 4r )= 120
5(2a1 + 4r) = 120
10a1 + 20r = 120   
a1 + 2r = 12 ***** (  1 )

S8 = 8( a1 + a8 )/2
90 = 8 ( a1 + a1 + 7r)/2
180 = 8 ( 2a1 + 7r )
16a1 + 56r = 180
4a1 + 14r =  45 *****  ( 2 )

 a1 +  2r =   12  ( vezes  - 4 )
4a1 + 14r = 45
-----------------------
-4a1 - 8r  = - 48
 4a1 + 14r = 45
-------------------------
//          6r  = -3
r =  -3/6 =  - 1/2 ***

a1 + 2r = 12
a1 + 2 ( - 1/2) = 12
a1 -  1 = 12
a1 = 12 + 1 = 13 ****

a1 + 6r =  13 + 6 ( - 1/2) = 13 - 3 = 10 ****** Resposta de sétimo termo

Answer 2

Dados:

$\displaystyle \mathsf{S_5 = 60}\\ \displaystyle \mathsf{S_8 = 90}\\ \displaystyle \mathsf{a_7 = \: ?}$

Cálculo:

$\displaystyle \mathsf{S_5}\\ \\ \displaystyle \mathsf{a1 = \: ?}\\ \displaystyle \mathsf{n = 5}\\ \displaystyle \mathsf{r = \: ?}\\ \displaystyle \mathsf{a5 = \: ?}\\ \displaystyle \mathsf{S_5 = 60}$

$\displaystyle \mathsf{a_n = a_1 + (n - 1).r}\\ \displaystyle \mathsf{a_5 = a_1 + (5 - 1).r}\\ \displaystyle \mathsf{a_5 = a_1 + 4r}$

$\displaystyle \mathsf{S_n = \frac{(a_1+a_n).n}{2} }\\ \\ \displaystyle \mathsf{S_5 = \frac{(a_1 + a_5).5}{2}}\\ \\ \displaystyle \mathsf{60 = \frac{(a_1 + a_1 + 4r).5}{2}}\\ \\ \displaystyle \mathsf{60 = \frac{(2a_1 + 4r).5}{2}} \\ \\ \displaystyle \mathsf{60.2 = (2a_1 + 4r).5} \\ \displaystyle \mathsf{(2a_1 + 4r).5 = 120}\\ \displaystyle \mathsf{2a_1 + 4r = \frac{120}{5}}\\ \displaystyle \mathsf{2a_1 + 4r = 24 \div 2}\\ \displaystyle \boxed{\mathsf{a_1 + 2r = 12}}$

$\displaystyle \mathsf{S_8}\\ \\ \displaystyle \mathsf{a_1 = \: ?}\\ \displaystyle \mathsf{n = \: 8}\\ \displaystyle \mathsf{r = \: ?}\\ \displaystyle \mathsf{a_8 = \: ?}\\ \displaystyle \mathsf{S_8 = 90 }$

$\displaystyle \mathsf{a_n = a_1 + (n - 1).r}\\ \displaystyle \mathsf{a_8 = a_1 + (8 - 1).r}\\ \displaystyle \mathsf{a_8 = a_1 + 7r}$

$\displaystyle \mathsf{S_n = \frac{(a_1 + a_n).n}{2}}\\ \\ \displaystyle \mathsf{S_8 = \frac{(a_1 + a_8).8}{2}}\\ \\ \displaystyle \mathsf{90 = \frac{(a_1 + a_1 + 7r).8}{2}}\\ \displaystyle \mathsf{4.(2a_1 + 7r) = 90}\\ \displaystyle \mathsf{8a_1 + 28r = 90 \div 2}\\ \displaystyle \mathsf{4a_1 + 14r = 45 }\\ \displaystyle \mathsf{}$
 
Sistema de equações:

$\displaystyle \mathsf{ \left \{ {{a_1 + 2r = 12} \atop {4a_1 + 14r = 45}} \right. }\\ \\ \\ \displaystyle \mathsf{a_1 = 12 - 2r }\\ \displaystyle \mathsf{4a_1 + 14r = 45}\\ \displaystyle \mathsf{4(12 - 2r) + 14r = 45}\\ \displaystyle \mathsf{48 - 8r + 14r = 45}\\ \displaystyle \mathsf{48 + 6r = 45}\\ \displaystyle \mathsf{6r = 45 - 48}\\ \displaystyle \mathsf{6r = -3}\\ \displaystyle \mathsf{r = \frac{-3}{6} = \frac{-1}{2}}$

$\displaystyle \mathsf{a_1 + 2r = 12}\\ \displaystyle \mathsf{a_1 + 2(\frac{-1}{2})= 12} \displaystyle \mathsf{a_1 + 2r = 12}\\ \displaystyle \mathsf{a_1 - 1= 12}\\ \displaystyle \mathsf{a_1 = 12 + 1}\\ \displaystyle \mathsf{a_1 = 13}$

$\displaystyle \mathsf{a_n = a_1 + (n - 1).r}\\ \displaystyle \mathsf{a_7 = 13 + (7 - 1).(\frac{-1}{2})}\\ \displaystyle \mathsf{a_7 = 13 - 3}\\ \displaystyle \mathsf{a_7 = 10}$