Question

Dada a PA com a6+a9=11 e a4+a13=102. Determine a20

Answer 1

P.A.(
a 6 + a 9 = 11

a 6 = a 1 + 5 r
a 9 = a 1 + 8 r
--------------------Somar
11 = 2 a 1 + 13 r
2 a 1 + 13 r = 11
2 a 1 = 11 - 13 r
a 1 = (11 - 13 r) / 2 (1)

a 4 + a 9 = 102

a 4 = a 1 + 3 r
a 13 = a 1 + 12 r
------------------------- Somar
102 = 2 a 1 + 15 r 
2 a 1  + 15 r = 102
2 a 1 = 102 - 15 r
a 1 = (102 - 15 r ) / 2 (2 )
igualar (10 e(2)
( 11- 13 r) / 2 = (102 - 15 r ) / 2   (eliminar o denominador)
11 - 13 r = 102 - 15 r
-13 r  + 15 r = 102 - 11
 2 r = 91
r = 91/2   
 a 1 = (102 - 15.r) / 2
a 1 = (102 - 15. 91/2 ) / 2
a 1 = (102 / 1  - 1365 / 2 ) / 2
a 1 = (204/2 - 1365/2) / 2
  a 1 = (-1161 / 2) /2
a 1 = -1161 / 2 x 1/2
a 1 = -1161 / 4

a 20 = a 1 + (n - 1 ).r
a 20 = -1161/4 + (20 - 1 )..91/2
a 20 = -1161 / 4 + 19 . 91/2
a 20 = -1161/4  + 1729 / 2
a 20 = -1161/4 + 3458 / 4
a 20 = 2297 / 4