Question

Encontrar el termino de estas sucesiones
 termino • 7, 13, 19, 25 = 7°
termino • 12, 20, 28, 36 = 6°
termino • 16, 21, 26, 31 = 9°
termino • 5, 14, 23, 32 = 8°
 termino • 3, 10, 17, 24 = 15°
termino • 4, 14, 24, 24 = 16°
termino • 1, 9, 18, 27 = 8°
termino • 53, 49, 45, 41 = 12°
termino • 69, 63, 57, 51 = 9°
termino • 42, 36, 30, 24 = 16°

porfa ayuda :D

Answer 1

veamos la formula es aₓ = a₁ +(n-1)r

1)    7,13,19 .....   termino 7°   
a₁ = 7
n =7
razón = (13-7) = 6 aplicando la formula del termino enésimo
 a₇ = 7 + (7-1) *6
 a₇ = 7+(6) *6
a₇ = 7+36
a₇ = 43

2)  12, 20,28,36.......   termino 6°
 a₁ = 12
n= 6 
r =20-12 = 8
a₆ = 12+( 6-1)8
a₆ = 12+40
a₆= 52

3) 16,21,26..... termino  9°
a₁ = 16
n= 9
r = 21-16=5
a₉ = 16+(9-1)5
a₉ = 16+40
a₉ = 56
 4)  5,14,2332..... termino 8°
a₁ = 5
n= 8
r = 14-5 = 9
a₈ = 5+(8-1)8
a₈ = 5+56
a₈ = 61
5) 3,10,17,24,...... termino 15°
a= 3
n = 15 
r = 10-3= 7
a₁₅ = 3+(15-1)7
a₁₅= 3+14*7
a₁₅ = 3+98
a₁₅ = 101
6) 4,14 24,34.........16°
a = 4
n= 16
r = 14-4 = 10
a₁₆ = 4+(16-!)10
a₁₆ = 4+15*10
a₁₆ = 154
7) 1,9,18,27,------- 8° lo podemos ordenar así
9(1,2,3-----) ⇒la sucesión a encontrar sería 1,2,3....luego lo multiplicamos *9
a₁ = 1
n = 8
r = 1
a ₈= 1+(8-1)8
a₈ = 1+56
a ₈= 57    entonces 9*a₈
 a₈ = 57*9
 a= 613
8) 53,49,45,41....  termino 12°
a₁₁₁ = 53
n = 12
r = 49-53 = -4
a₁₂ = 53+(12-1)*-4
a₁₂ = 53+11*-4
a₁₂ = 53-44
a₁₂ = 11
9) 69,63,57,51......9°
a₁=69
n =9
r = 63-69 = -6
a₉ = 69+(9-1)*-6
a₉= 69 +8*-6
a₉= 69-48
a₉= 21
10) 42,36,30,24......... 16°
a₁= 42
r= 36-42= -6
n= 16
a = 42+(16-1)*-6
a = 42+15*-6
a = 42-90
a = 48