Uma progressão aritmética (abreviadamente, P. A. ) ou sequência aritmética é uma sequência numérica em que cada termo, a partir do segundo, é igual à soma do termo anterior com uma constante
r
{\displaystyle r}
. O número
r
{\displaystyle r}
é chamado de razão ou diferença comum da progressão aritmética.
Uma parte finita de uma progressão aritmética é chamada uma progressão aritmética finita ou apenas progressão aritmética. A soma de uma progressão aritmética finita é chamada uma série aritmética finita.
Notamos que, de modo geral, uma sequência numérica
(
a
n
)
n
∈
R
a
n
g
e
r
=
(
…
,
a
1
,
a
2
,
a
3
,
a
4
,
…
,
a
n
−
2
,
a
n
−
1
,
a
n
,
…
)
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n})}}}=(\ldots ,~a_{1},~a_{2},~a_{3},~a_{4},~\ldots ,~a_{n-2},~a_{n-1},~a_{n},~\ldots )}
é uma P.A. quando definida recursivamente por:
…
{\displaystyle \qquad \ldots }
a
2
=
a
1
+
r
⇒
r
=
a
2
−
a
1
{\displaystyle \qquad a_{2}=a_{1}+r\quad \Rightarrow \quad r=a_{2}-a_{1}}
a
3
=
a
2
+
r
⇒
r
=
a
3
−
a
2
{\displaystyle \qquad a_{3}=a_{2}+r\quad \Rightarrow \quad r=a_{3}-a_{2}}
a
4
=
a
3
+
r
⇒
r
=
a
4
−
a
3
{\displaystyle \qquad a_{4}=a_{3}+r\quad \Rightarrow \quad r=a_{4}-a_{3}}
…
{\displaystyle \qquad \ldots }
a
n
−
2
=
a
n
−
3
+
r
⇒
r
=
a
n
−
2
−
a
n
−
3
{\displaystyle \qquad a_{n-2}=a_{n-3}+r\quad \Rightarrow \quad r=a_{n-2}-a_{n-3}}
a
n
−
1
=
a
n
−
2
+
r
⇒
r
=
a
n
−
1
−
a
n
−
2
{\displaystyle \qquad a_{n-1}=a_{n-2}+r\quad \Rightarrow \quad r=a_{n-1}-a_{n-2}}
a
n
=
a
n
−
1
+
r
⇒
r
=
a
n
−
a
n
−
1
{\displaystyle \qquad a_{n}=a_{n-1}+r\quad \Rightarrow \quad r=a_{n}-a_{n-1}}
…
{\displaystyle \qquad \ldots }
Comparando, temos:
r
=
…
=
a
2
−
a
1
=
a
3
−
a
2
=
a
4
−
a
3
=
…
=
a
n
−
1
−
a
n
−
2
=
a
n
−
a
n
−
1
=
…
{\displaystyle r=\ldots =a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=\ldots =a_{n-1}-a_{n-2}=a_{n}-a_{n-1}=\ldots }
onde
a
n
{\displaystyle a_{n}}
é o n -ésino termo da sequência
(
a
n
)
{\displaystyle (a_{n})}
n
{\displaystyle n}
corresponde ao número de termos
−
{\displaystyle -}
até
a
n
{\displaystyle a_{n}}
o primeiro termo,
a
1
{\displaystyle a_{1}}
é um número dado
o número
r
{\displaystyle r}
é chamado de razão da progressão aritmética
Alguns exemplos de progressões aritméticas:
(
a
n
)
n
=
1
:
1
:
∞
r
=
3
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n})}}}=(1,~4,~7,~10,~13,~\ldots )}
é uma progressão aritmética em que o primeiro termo
a
1
{\displaystyle a_{1}}
é igual a
1
{\displaystyle 1}
e a razão
r
{\displaystyle r}
é igual a
3
{\displaystyle 3}
(
a
n
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n})}}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
é uma P.A. em que
a
1
=
−
2
{\displaystyle a_{1}=-2}
e
r
=
−
2
{\displaystyle r=-2}
(
a
n
)
n
=
1
:
1
:
∞
r
=
0
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n})}}}=(6,~6,~6,~6,~6,~\ldots )}
é uma P.A. com
a
1
=
6
{\displaystyle a_{1}=6}
e
r
=
0
{\displaystyle r=0}
Fórmulas para
a
n
{\displaystyle a_{n}}
,
a
n
−
p
{\displaystyle a_{n-p}}
,
a
n
+
p
{\displaystyle a_{n+p}}
editar
a
n
=
a
m
+
(
n
−
m
)
⋅
r
{\displaystyle a_{n}=a_{m}+(n-m)\cdot r}
editar
O n -ésimo termo de uma progressão aritmética, denotado por
a
n
{\displaystyle a_{n}}
, pode ser obtido por meio da fórmula:
a
n
=
a
m
+
(
n
−
m
)
⋅
r
{\displaystyle a_{n}=a_{m}+(n-m)\cdot r}
portanto uma sequência
(
a
n
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n})}}}}
também pode ser expressa por
(
a
n
=
a
m
+
(
n
−
m
)
⋅
r
)
n
∈
R
a
n
g
e
=
(
a
m
+
(
n
−
m
)
⋅
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n}=a_{m}+(n-m)\cdot r)}}={\underset {n~\in ~Range}{(a_{m}+(n-m)\cdot r)}}}
a
n
=
a
1
+
(
n
−
1
)
⋅
r
{\displaystyle a_{n}=a_{1}+(n-1)\cdot r}
editar
ou pela fórmula:
a
n
=
a
1
+
(
n
−
1
)
⋅
r
{\displaystyle a_{n}=a_{1}+(n-1)\cdot r}
de forma semelhante, uma sequência
(
a
n
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n})}}}}
também pode ser expressa por
(
a
n
=
a
1
+
(
n
−
1
)
⋅
r
)
n
∈
R
a
n
g
e
=
(
a
1
+
(
n
−
1
)
⋅
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n}=a_{1}+(n-1)\cdot r)}}={\underset {n~\in ~Range}{(a_{1}+(n-1)\cdot r)}}}
em que:
a
n
{\displaystyle a_{n}}
é o n -ésimo termo da sequência
(
a
n
)
{\displaystyle (a_{n})}
n
{\displaystyle n}
é o número de termos
−
{\displaystyle -}
até
a
n
{\displaystyle a_{n}}
a
1
{\displaystyle a_{1}}
é o primeiro termo
r
{\displaystyle r}
é a razão
(
a
n
)
n
=
1
:
1
:
∞
r
=
3
=
(
a
n
=
a
m
+
(
n
−
m
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
1
r
=
3
=
(
a
n
=
a
1
+
(
n
−
1
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
a
n
=
1
+
(
n
−
1
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
a
n
=
1
+
3
n
−
3
)
n
=
1
:
1
:
∞
=
(
a
n
=
3
n
−
2
)
n
=
1
:
1
:
∞
=
(
3
n
−
2
)
n
=
1
:
1
:
∞
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n})}}}={\overset {m~=~1\quad r~=~3}{\underset {n~=~1:1:\infty }{(a_{n}=a_{m}+(n-m)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{n}=a_{1}+(n-1)\cdot 3)}}={\underset {n~=~1:1:\infty }{(a_{n}=1+(n-1)\cdot 3)}}={\underset {n~=~1:1:\infty }{(a_{n}=1+3n-3)}}={\underset {n~=~1:1:\infty }{(a_{n}=3n-2)}}={\underset {n~=~1:1:\infty }{(3n-2)}}=(1,~4,~7,~10,~13,~\ldots )}
(
a
n
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
a
n
=
a
m
+
(
n
−
m
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
3
r
=
−
2
=
(
a
n
=
a
3
+
(
n
−
3
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
a
n
=
−
6
+
(
n
−
3
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
a
n
=
−
6
−
2
n
+
6
)
n
=
1
:
1
:
∞
=
(
a
n
=
−
2
n
)
n
=
1
:
1
:
∞
=
(
−
2
n
)
n
=
1
:
1
:
∞
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n})}}}={\overset {m~=~3\quad r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n}=a_{m}+(n-m)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{n}=a_{3}+(n-3)\cdot -2)}}={\underset {n~=~1:1:\infty }{(a_{n}=-6+(n-3)\cdot -2)}}={\underset {n~=~1:1:\infty }{(a_{n}=-6-2n+6)}}={\underset {n~=~1:1:\infty }{(a_{n}=-2n)}}={\underset {n~=~1:1:\infty }{(-2n)}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
(
a
n
)
n
=
1
:
1
:
∞
r
=
0
=
(
a
n
=
a
m
+
(
n
−
m
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
5
r
=
0
=
(
a
n
=
a
5
+
(
n
−
5
)
⋅
0
)
n
=
1
:
1
:
∞
=
(
6
)
n
=
1
:
1
:
∞
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n})}}}={\overset {m~=~5\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{n}=a_{m}+(n-m)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{n}=a_{5}+(n-5)\cdot 0)}}={\underset {n~=~1:1:\infty }{(6)}}=(6,~6,~6,~6,~6,~\ldots )}
a
n
−
p
=
a
m
+
(
n
−
m
−
p
)
⋅
r
{\displaystyle a_{n-p}=a_{m}+(n-m-p)\cdot r}
editar
Outras fórmulas para P.A. são:
a
n
−
p
=
a
m
+
(
n
−
m
−
p
)
⋅
r
{\displaystyle a_{n-p}=a_{m}+(n-m-p)\cdot r}
onde a sequência
(
a
n
−
p
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n-p})}}}}
pode ser expressa por
(
a
n
−
p
=
a
m
+
(
n
−
m
−
p
)
⋅
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n-p}=a_{m}+(n-m-p)\cdot r)}}}
(
a
n
−
p
)
n
=
2
:
1
:
∞
p
=
1
r
=
3
=
(
a
n
−
p
=
a
m
+
(
n
−
m
−
p
)
⋅
r
)
n
=
2
:
1
:
∞
m
=
2
p
=
1
r
=
3
=
(
a
n
−
1
=
a
2
+
(
n
−
2
−
1
)
⋅
3
)
n
=
2
:
1
:
∞
=
(
a
n
−
1
=
4
+
(
n
−
3
)
⋅
3
)
n
=
2
:
1
:
∞
=
(
a
n
−
1
=
4
+
3
n
−
9
)
n
=
2
:
1
:
∞
=
(
a
n
−
1
=
3
n
−
5
)
n
=
2
:
1
:
∞
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {p~=~1\quad r~=~3}{\underset {n~=~2:1:\infty }{(a_{n-p})}}}={\overset {m~=~2\quad p~=~1\quad r~=~3}{\underset {n~=~2:1:\infty }{(a_{n-p}=a_{m}+(n-m-p)\cdot r)}}}={\underset {n~=~2:1:\infty }{(a_{n-1}=a_{2}+(n-2-1)\cdot 3)}}={\underset {n~=~2:1:\infty }{(a_{n-1}=4+(n-3)\cdot 3)}}={\underset {n~=~2:1:\infty }{(a_{n-1}=4+3n-9)}}={\underset {n~=~2:1:\infty }{(a_{n-1}=3n-5)}}=(1,~4,~7,~10,~13,~\ldots )}
(
a
n
−
p
)
n
=
6
:
1
:
∞
p
=
5
r
=
−
2
=
(
a
n
−
p
=
a
m
+
(
n
−
m
−
p
)
⋅
r
)
n
=
6
:
1
:
∞
m
=
4
p
=
5
r
=
−
2
=
(
a
n
−
5
=
a
4
+
(
n
−
4
−
5
)
⋅
−
2
)
n
=
6
:
1
:
∞
=
(
a
n
−
5
=
−
8
+
(
n
−
9
)
⋅
−
2
)
n
=
6
:
1
:
∞
=
(
a
n
−
5
=
−
8
−
2
n
+
18
)
n
=
6
:
1
:
∞
=
(
a
n
−
5
=
−
2
n
+
10
)
n
=
1
:
1
:
∞
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {p~=~5\quad r~=~-2}{\underset {n~=~6:1:\infty }{(a_{n-p})}}}={\overset {m~=~4\quad p~=~5\quad r~=~-2}{\underset {n~=~6:1:\infty }{(a_{n-p}=a_{m}+(n-m-p)\cdot r)}}}={\underset {n~=~6:1:\infty }{(a_{n-5}=a_{4}+(n-4-5)\cdot -2)}}={\underset {n~=~6:1:\infty }{(a_{n-5}=-8+(n-9)\cdot -2)}}={\underset {n~=~6:1:\infty }{(a_{n-5}=-8-2n+18)}}={\underset {n~=~1:1:\infty }{(a_{n-5}=-2n+10)}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
(
a
n
−
p
)
n
=
1
:
1
:
∞
p
=
6
r
=
0
=
(
a
n
−
p
=
a
m
+
(
n
−
m
−
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
9
p
=
6
r
=
0
=
(
a
n
−
6
=
a
9
+
(
n
−
9
−
6
)
⋅
0
)
n
=
1
:
1
:
∞
=
(
6
)
n
=
1
:
1
:
∞
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {p~=~6\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{n-p})}}}={\overset {m~=~9\quad p~=~6\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{n-p}=a_{m}+(n-m-p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{n-6}=a_{9}+(n-9-6)\cdot 0)}}={\underset {n~=~1:1:\infty }{(6)}}=(6,~6,~6,~6,~6,~\ldots )}
a
n
+
p
=
a
m
+
(
n
−
m
+
p
)
⋅
r
{\displaystyle a_{n+p}=a_{m}+(n-m+p)\cdot r}
editar
a
n
+
p
=
a
m
+
(
n
−
m
+
p
)
⋅
r
{\displaystyle a_{n+p}=a_{m}+(n-m+p)\cdot r}
em que a sequência
(
a
n
+
p
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n+p})}}}}
é expressa por
(
a
n
+
p
=
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n+p}=a_{m}+(n-m+p)\cdot r)}}}
(
a
n
+
p
)
n
=
−
2
:
1
:
∞
p
=
3
r
=
3
=
(
a
n
+
p
=
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
−
2
:
1
:
∞
m
=
2
p
=
3
r
=
3
=
(
a
n
+
3
=
a
2
+
(
n
−
2
+
3
)
⋅
3
)
n
=
−
2
:
1
:
∞
=
(
a
n
+
3
=
4
+
(
n
+
1
)
⋅
3
)
n
=
−
2
:
1
:
∞
=
(
a
n
+
3
=
4
+
3
n
+
3
)
n
=
−
2
:
1
:
∞
=
(
a
n
+
3
=
3
n
+
7
)
n
=
−
2
:
1
:
∞
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {p~=~3\quad r~=~3}{\underset {n~=~-2:1:\infty }{(a_{n+p})}}}={\overset {m~=~2\quad p~=~3\quad r~=~3}{\underset {n~=~-2:1:\infty }{(a_{n+p}=a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~-2:1:\infty }{(a_{n+3}=a_{2}+(n-2+3)\cdot 3)}}={\underset {n~=~-2:1:\infty }{(a_{n+3}=4+(n+1)\cdot 3)}}={\underset {n~=~-2:1:\infty }{(a_{n+3}=4+3n+3)}}={\underset {n~=~-2:1:\infty }{(a_{n+3}=3n+7)}}=(1,~4,~7,~10,~13,~\ldots )}
(
a
n
+
p
)
n
=
−
1
:
1
:
∞
p
=
2
r
=
−
2
=
(
a
n
+
p
=
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
−
1
:
1
:
∞
m
=
3
p
=
2
r
=
−
2
=
(
a
n
+
2
=
a
3
+
(
n
−
3
+
2
)
⋅
−
2
)
n
=
−
1
:
1
:
∞
=
(
a
n
+
2
=
−
6
+
(
n
−
1
)
⋅
−
2
)
n
=
−
1
:
1
:
∞
=
(
a
n
+
2
=
−
6
−
2
n
+
2
)
n
=
−
1
:
1
:
∞
=
(
a
n
+
2
=
−
2
n
−
4
)
n
=
−
1
:
1
:
∞
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {p~=~2\quad r~=~-2}{\underset {n~=~-1:1:\infty }{(a_{n+p})}}}={\overset {m~=~3\quad p~=~2\quad r~=~-2}{\underset {n~=~-1:1:\infty }{(a_{n+p}=a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~-1:1:\infty }{(a_{n+2}=a_{3}+(n-3+2)\cdot -2)}}={\underset {n~=~-1:1:\infty }{(a_{n+2}=-6+(n-1)\cdot -2)}}={\underset {n~=~-1:1:\infty }{(a_{n+2}=-6-2n+2)}}={\underset {n~=~-1:1:\infty }{(a_{n+2}=-2n-4)}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
p
=
1
r
=
0
=
(
a
n
+
p
=
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
4
p
=
1
r
=
0
=
(
a
n
+
1
=
a
4
+
(
n
−
4
+
1
)
⋅
0
)
n
=
1
:
1
:
∞
=
(
6
)
n
=
1
:
1
:
∞
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {p~=~1\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~4\quad p~=~1\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+p}=a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{n+1}=a_{4}+(n-4+1)\cdot 0)}}={\underset {n~=~1:1:\infty }{(6)}}=(6,~6,~6,~6,~6,~\ldots )}
a
n
−
p
=
a
n
−
s
−
(
p
−
s
)
⋅
r
{\displaystyle a_{n-p}=a_{n-s}-(p-s)\cdot r}
editar
a
n
−
p
=
a
n
−
s
−
(
p
−
s
)
⋅
r
{\displaystyle a_{n-p}=a_{n-s}-(p-s)\cdot r}
em que a sequência
(
a
n
−
p
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n-p})}}}}
é expressa por
(
a
n
−
s
−
(
p
−
s
)
⋅
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n-s}-(p-s)\cdot r)}}}
(
a
n
−
p
=
a
n
−
s
−
(
p
−
s
)
⋅
r
)
n
=
5
:
1
:
∞
p
=
4
s
=
7
r
=
3
=
(
a
n
−
4
=
a
n
−
7
−
(
4
−
7
)
⋅
3
)
n
=
5
:
1
:
∞
=
(
a
n
−
4
=
a
n
−
7
−
(
−
3
)
⋅
3
)
n
=
5
:
1
:
∞
=
(
a
n
−
4
=
a
n
−
7
+
9
)
n
=
5
:
1
:
∞
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {p~=~4\quad s~=~7\quad r~=~3}{\underset {n~=~5:1:\infty }{(a_{n-p}=a_{n-s}-(p-s)\cdot r)}}}={\underset {n~=~5:1:\infty }{(a_{n-4}=a_{n-7}-(4-7)\cdot 3)}}={\underset {n~=~5:1:\infty }{(a_{n-4}=a_{n-7}-(-3)\cdot 3)}}={\underset {n~=~5:1:\infty }{(a_{n-4}=a_{n-7}+9)}}=(1,~4,~7,~10,~13,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
3
p
=
2
r
=
−
2
=
(
a
3
+
(
n
−
3
+
2
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
+
(
n
−
1
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
−
2
n
+
2
)
n
=
1
:
1
:
∞
=
(
a
n
+
2
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
−
2
n
−
4
)
n
=
1
:
1
:
∞
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~3\quad p~=~2\quad r~=~-2}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{3}+(n-3+2)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6+(n-1)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6-2n+2)}}={\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+2})}}}={\underset {n~=~1:1:\infty }{(-2n-4)}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
0
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
4
p
=
1
r
=
0
=
(
a
4
+
(
n
−
4
+
1
)
⋅
0
)
n
=
1
:
1
:
∞
=
(
a
n
+
1
)
n
=
1
:
1
:
∞
r
=
0
=
(
6
)
n
=
1
:
1
:
∞
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~4\quad p~=~1\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{4}+(n-4+1)\cdot 0)}}={\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+1})}}}={\underset {n~=~1:1:\infty }{(6)}}=(6,~6,~6,~6,~6,~\ldots )}
a
n
+
p
=
a
n
−
q
+
(
p
+
q
)
r
{\displaystyle a_{n+p}=a_{n-q}+(p+q)r}
editar
a
n
+
p
=
a
n
−
q
+
(
p
+
q
)
r
{\displaystyle a_{n+p}=a_{n-q}+(p+q)r}
em que a sequência
(
a
n
+
p
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n+p})}}}}
é expressa por
(
a
n
−
q
+
(
p
+
q
)
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n-q}+(p+q)r)}}}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
3
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
2
p
=
3
r
=
3
=
(
a
2
+
(
n
−
2
+
3
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
(
n
+
1
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
3
n
+
3
)
n
=
1
:
1
:
∞
=
(
a
n
+
3
)
n
=
1
:
1
:
∞
r
=
3
=
(
3
n
+
7
)
n
=
1
:
1
:
∞
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~2\quad p~=~3\quad r~=~3}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{2}+(n-2+3)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+(n+1)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+3n+3)}}={\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+3})}}}={\underset {n~=~1:1:\infty }{(3n+7)}}=(1,~4,~7,~10,~13,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
3
p
=
2
r
=
−
2
=
(
a
3
+
(
n
−
3
+
2
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
+
(
n
−
1
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
−
2
n
+
2
)
n
=
1
:
1
:
∞
=
(
a
n
+
2
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
−
2
n
−
4
)
n
=
1
:
1
:
∞
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~3\quad p~=~2\quad r~=~-2}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{3}+(n-3+2)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6+(n-1)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6-2n+2)}}={\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+2})}}}={\underset {n~=~1:1:\infty }{(-2n-4)}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
0
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
4
p
=
1
r
=
0
=
(
a
4
+
(
n
−
4
+
1
)
⋅
0
)
n
=
1
:
1
:
∞
=
(
a
n
+
1
)
n
=
1
:
1
:
∞
r
=
0
=
(
6
)
n
=
1
:
1
:
∞
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~4\quad p~=~1\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{4}+(n-4+1)\cdot 0)}}={\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+1})}}}={\underset {n~=~1:1:\infty }{(6)}}=(6,~6,~6,~6,~6,~\ldots )}
a
n
−
p
=
a
n
+
q
−
(
p
+
q
)
r
{\displaystyle a_{n-p}=a_{n+q}-(p+q)r}
editar
a
n
−
p
=
a
n
+
q
−
(
p
+
q
)
r
{\displaystyle a_{n-p}=a_{n+q}-(p+q)r}
em que a sequência
(
a
n
−
p
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n-p})}}}}
é expressa por
(
a
n
+
q
−
(
p
+
q
)
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n+q}-(p+q)r)}}}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
3
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
2
p
=
3
r
=
3
=
(
a
2
+
(
n
−
2
+
3
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
(
n
+
1
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
3
n
+
3
)
n
=
1
:
1
:
∞
=
(
a
n
+
3
)
n
=
1
:
1
:
∞
r
=
3
=
(
3
n
+
7
)
n
=
1
:
1
:
∞
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~2\quad p~=~3\quad r~=~3}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{2}+(n-2+3)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+(n+1)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+3n+3)}}={\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+3})}}}={\underset {n~=~1:1:\infty }{(3n+7)}}=(1,~4,~7,~10,~13,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
3
p
=
2
r
=
−
2
=
(
a
3
+
(
n
−
3
+
2
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
+
(
n
−
1
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
−
2
n
+
2
)
n
=
1
:
1
:
∞
=
(
a
n
+
2
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
−
2
n
−
4
)
n
=
1
:
1
:
∞
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~3\quad p~=~2\quad r~=~-2}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{3}+(n-3+2)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6+(n-1)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6-2n+2)}}={\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+2})}}}={\underset {n~=~1:1:\infty }{(-2n-4)}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
0
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
4
p
=
1
r
=
0
=
(
a
4
+
(
n
−
4
+
1
)
⋅
0
)
n
=
1
:
1
:
∞
=
(
a
n
+
1
)
n
=
1
:
1
:
∞
r
=
0
=
(
6
)
n
=
1
:
1
:
∞
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~4\quad p~=~1\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{4}+(n-4+1)\cdot 0)}}={\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+1})}}}={\underset {n~=~1:1:\infty }{(6)}}=(6,~6,~6,~6,~6,~\ldots )}
a
n
+
p
=
a
n
+
q
+
(
p
−
q
)
r
{\displaystyle a_{n+p}=a_{n+q}+(p-q)r}
editar
a
n
+
p
=
a
n
+
q
+
(
p
−
q
)
r
{\displaystyle a_{n+p}=a_{n+q}+(p-q)r}
em que a sequência
(
a
n
+
p
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n+p})}}}}
é expressa por
(
a
n
+
q
+
(
p
−
q
)
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n+q}+(p-q)r)}}}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
3
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
2
p
=
3
r
=
3
=
(
a
2
+
(
n
−
2
+
3
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
(
n
+
1
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
3
n
+
3
)
n
=
1
:
1
:
∞
=
(
a
n
+
3
)
n
=
1
:
1
:
∞
r
=
3
=
(
3
n
+
7
)
n
=
1
:
1
:
∞
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~2\quad p~=~3\quad r~=~3}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{2}+(n-2+3)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+(n+1)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+3n+3)}}={\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+3})}}}={\underset {n~=~1:1:\infty }{(3n+7)}}=(1,~4,~7,~10,~13,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
3
p
=
2
r
=
−
2
=
(
a
3
+
(
n
−
3
+
2
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
+
(
n
−
1
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
−
2
n
+
2
)
n
=
1
:
1
:
∞
=
(
a
n
+
2
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
−
2
n
−
4
)
n
=
1
:
1
:
∞
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~3\quad p~=~2\quad r~=~-2}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{3}+(n-3+2)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6+(n-1)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6-2n+2)}}={\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+2})}}}={\underset {n~=~1:1:\infty }{(-2n-4)}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
0
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
4
p
=
1
r
=
0
=
(
a
4
+
(
n
−
4
+
1
)
⋅
0
)
n
=
1
:
1
:
∞
=
(
a
n
+
1
)
n
=
1
:
1
:
∞
r
=
0
=
(
6
)
n
=
1
:
1
:
∞
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~4\quad p~=~1\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{4}+(n-4+1)\cdot 0)}}={\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+1})}}}={\underset {n~=~1:1:\infty }{(6)}}=(6,~6,~6,~6,~6,~\ldots )}
a
n
−
p
=
a
n
−
p
r
{\displaystyle a_{n-p}=a_{n}-pr}
editar
a
n
−
p
=
a
n
−
p
r
{\displaystyle a_{n-p}=a_{n}-pr}
em que a sequência
(
a
n
−
p
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n-p})}}}}
é expressa por
(
a
n
−
p
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n}-pr)}}}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
3
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
2
p
=
3
r
=
3
=
(
a
2
+
(
n
−
2
+
3
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
(
n
+
1
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
3
n
+
3
)
n
=
1
:
1
:
∞
=
(
a
n
+
3
)
n
=
1
:
1
:
∞
r
=
3
=
(
3
n
+
7
)
n
=
1
:
1
:
∞
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~2\quad p~=~3\quad r~=~3}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{2}+(n-2+3)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+(n+1)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+3n+3)}}={\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+3})}}}={\underset {n~=~1:1:\infty }{(3n+7)}}=(1,~4,~7,~10,~13,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
3
p
=
2
r
=
−
2
=
(
a
3
+
(
n
−
3
+
2
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
+
(
n
−
1
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
−
2
n
+
2
)
n
=
1
:
1
:
∞
=
(
a
n
+
2
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
−
2
n
−
4
)
n
=
1
:
1
:
∞
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~3\quad p~=~2\quad r~=~-2}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{3}+(n-3+2)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6+(n-1)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6-2n+2)}}={\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+2})}}}={\underset {n~=~1:1:\infty }{(-2n-4)}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
0
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
4
p
=
1
r
=
0
=
(
a
4
+
(
n
−
4
+
1
)
⋅
0
)
n
=
1
:
1
:
∞
=
(
a
n
+
1
)
n
=
1
:
1
:
∞
r
=
0
=
(
6
)
n
=
1
:
1
:
∞
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~4\quad p~=~1\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{4}+(n-4+1)\cdot 0)}}={\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+1})}}}={\underset {n~=~1:1:\infty }{(6)}}=(6,~6,~6,~6,~6,~\ldots )}
a
n
+
p
=
a
n
+
p
r
{\displaystyle a_{n+p}=a_{n}+pr}
editar
a
n
+
p
=
a
n
+
p
r
{\displaystyle a_{n+p}=a_{n}+pr}
em que a sequência
(
a
n
+
p
)
n
∈
R
a
n
g
e
r
{\displaystyle {\overset {r}{\underset {n~\in ~Range}{(a_{n+p})}}}}
é expressa por
(
a
n
+
p
r
)
n
∈
R
a
n
g
e
{\displaystyle {\underset {n~\in ~Range}{(a_{n}+pr)}}}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
3
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
2
p
=
3
r
=
3
=
(
a
2
+
(
n
−
2
+
3
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
(
n
+
1
)
⋅
3
)
n
=
1
:
1
:
∞
=
(
4
+
3
n
+
3
)
n
=
1
:
1
:
∞
=
(
a
n
+
3
)
n
=
1
:
1
:
∞
r
=
3
=
(
3
n
+
7
)
n
=
1
:
1
:
∞
=
(
1
,
4
,
7
,
10
,
13
,
…
)
{\displaystyle {\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~2\quad p~=~3\quad r~=~3}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{2}+(n-2+3)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+(n+1)\cdot 3)}}={\underset {n~=~1:1:\infty }{(4+3n+3)}}={\overset {r~=~3}{\underset {n~=~1:1:\infty }{(a_{n+3})}}}={\underset {n~=~1:1:\infty }{(3n+7)}}=(1,~4,~7,~10,~13,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
3
p
=
2
r
=
−
2
=
(
a
3
+
(
n
−
3
+
2
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
+
(
n
−
1
)
⋅
−
2
)
n
=
1
:
1
:
∞
=
(
−
6
−
2
n
+
2
)
n
=
1
:
1
:
∞
=
(
a
n
+
2
)
n
=
1
:
1
:
∞
r
=
−
2
=
(
−
2
n
−
4
)
n
=
1
:
1
:
∞
=
(
−
2
,
−
4
,
−
6
,
−
8
,
−
10
,
…
)
{\displaystyle {\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~3\quad p~=~2\quad r~=~-2}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{3}+(n-3+2)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6+(n-1)\cdot -2)}}={\underset {n~=~1:1:\infty }{(-6-2n+2)}}={\overset {r~=~-2}{\underset {n~=~1:1:\infty }{(a_{n+2})}}}={\underset {n~=~1:1:\infty }{(-2n-4)}}=(-2,~-4,~-6,~-8,~-10,~\ldots )}
(
a
n
+
p
)
n
=
1
:
1
:
∞
r
=
0
=
(
a
m
+
(
n
−
m
+
p
)
⋅
r
)
n
=
1
:
1
:
∞
m
=
4
p
=
1
r
=
0
=
(
a
4
+
(
n
−
4
+
1
)
⋅
0
)
n
=
1
:
1
:
∞
=
(
a
n
+
1
)
n
=
1
:
1
:
∞
r
=
0
=
(
6
)
n
=
1
:
1
:
∞
=
(
6
,
6
,
6
,
6
,
6
,
…
)
{\displaystyle {\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+p})}}}={\overset {m~=~4\quad p~=~1\quad r~=~0}{\underset {n~=~1:1:\infty }{(a_{m}+(n-m+p)\cdot r)}}}={\underset {n~=~1:1:\infty }{(a_{4}+(n-4+1)\cdot 0)}}={\overset {r~=~0}{\underset {n~=~1:1:\infty }{(a_{n+1})}}}={\underset {n~=~1:1:\infty }{(6)}}=(6,~6,~6,~6,~6,~\ldots )}
Demonstrações para
a
n
{\displaystyle a_{n}}
,
a
n
−
p
{\displaystyle a_{n-p}}
,
a
n
+
p
{\displaystyle a_{n+p}}
editar
a
n
=
a
m
+
(
n
−
m
)
⋅
r
{\displaystyle a_{n}=a_{m}+(n-m)\cdot r}
editar
a
n
−
p
=
a
m
+
(
n
−
m
−
p
)
⋅
r
{\displaystyle a_{n-p}=a_{m}+(n-m-p)\cdot r}
editar
a
n
+
p
=
a
m
+
(
n
−
m
+
p
)
⋅
r
{\displaystyle a_{n+p}=a_{m}+(n-m+p)\cdot r}
editar
Tomando como referência índices quaisquer, por exemplo 2 e 3:
(
…
,
a
2
−
4
r
⏟
a
−
2
,
a
2
−
3
r
⏟
a
−
1
,
a
2
−
2
r
⏟
a
0
,
a
2
−
r
⏟
a
1
,
a
2
,
a
2
+
r
⏟
a
3
,
a
2
+
2
r
⏟
a
4
,
a
2
+
3
r
⏟
a
5
,
…
,
a
2
+
(
n
−
4
)
⋅
r
⏟
a
n
−
2
,
a
2
+
(
n
−
3
)
⋅
r
⏟
a
n
−
1
,
a
2
+
(
n
−
2
)
⋅
r
⏟
a
n
,
a
2
+
(
n
−
1
)
⋅
r
⏟
a
n
+
1
,
a
2
+
(
n
−
0
)
⋅
r
⏟
a
n
+
2
,
…
)
{\displaystyle \qquad (\ldots ,\quad \underbrace {a_{2}-4r} _{a_{-2}},\quad \underbrace {a_{2}-3r} _{a_{-1}},\quad \underbrace {a_{2}-2r} _{a_{0}},\quad \underbrace {a_{2}-r} _{a_{1}},\quad a_{2},\quad \underbrace {a_{2}+r} _{a_{3}},\quad \underbrace {a_{2}+2r} _{a_{4}},\quad \underbrace {a_{2}+3r} _{a_{5}},\quad \ldots ,\quad \underbrace {a_{2}+(n-4)\cdot r} _{a_{n-2}},\quad \underbrace {a_{2}+(n-3)\cdot r} _{a_{n-1}},\quad \underbrace {a_{2}+(n-2)\cdot r} _{a_{n}},\quad \underbrace {a_{2}+(n-1)\cdot r} _{a_{n+1}},\quad \underbrace {a_{2}+(n-0)\cdot r} _{a_{n+2}},\quad \ldots )}
(
…
,
a
3
−
5
r
⏟
a
−
2
,
a
3
−
4
r
⏟
a
−
1
,
a
3
−
3
r
⏟
a
0
,
a
3
−
2
r
⏟
a
1
,
a
3
−
r
⏟
a
2
,
a
3
,
a
3
+
r
⏟
a
4
,
a
3
+
2
r
⏟
a
5
,
a
3
+
3
r
⏟
a
6
,
…
,
a
3
+
(
n
−
5
)
⋅
r
⏟
a
n
−
2
,
a
3
+
(
n
−
4
)
⋅
r
⏟
a
n
−
1
,
a
3
+
(
n
−
3
)
⋅
r
⏟
a
n
,
a
3
+
(
n
−
2
)
⋅
r
⏟
a
n
+
1
,
a
3
+
(
n
−
1
)
⋅
r
⏟
a
n
+
2
,
…
)
{\displaystyle \qquad (\ldots ,\quad \underbrace {a_{3}-5r} _{a_{-2}},\quad \underbrace {a_{3}-4r} _{a_{-1}},\quad \underbrace {a_{3}-3r} _{a_{0}},\quad \underbrace {a_{3}-2r} _{a_{1}},\quad \underbrace {a_{3}-r} _{a_{2}},\quad a_{3},\quad \underbrace {a_{3}+r} _{a_{4}},\quad \underbrace {a_{3}+2r} _{a_{5}},\quad \underbrace {a_{3}+3r} _{a_{6}},\quad \ldots ,\quad \underbrace {a_{3}+(n-5)\cdot r} _{a_{n-2}},\quad \underbrace {a_{3}+(n-4)\cdot r} _{a_{n-1}},\quad \underbrace {a_{3}+(n-3)\cdot r} _{a_{n}},\quad \underbrace {a_{3}+(n-2)\cdot r} _{a_{n+1}},\quad \underbrace {a_{3}+(n-1)\cdot r} _{a_{n+2}},\quad \ldots )}
observe que qualquer que seja o índice tomado como referência, são sempre válidas e dedutíveis as fórmulas:
a
n
=
a
m
+
(
n
−
m
)
⋅
r
{\displaystyle a_{n}=a_{m}+(n-m)\cdot r}
ao passar de
a
m
{\displaystyle a_{m}}
para
a
n
{\displaystyle a_{n}}
, avançamos
(
n
−
m
)
{\displaystyle (n-m)}
termos, ou seja, basta somar
(
n
−
m
)
{\displaystyle (n-m)}
vezes a razão ao
m
{\displaystyle m}
º termo
Note que:
a
9
=
a
4
+
5
r
{\displaystyle a_{9}=a_{4}+5r}
, pois, ao passar de
a
4
{\displaystyle a_{4}}
para
a
9
{\displaystyle a_{9}}
, avançamos
5
{\displaystyle 5}
termos
a
3
=
a
15
−
12
r
{\displaystyle a_{3}=a_{15}-12r}
, pois retrocedemos 12 termos ao passar de
a
15
{\displaystyle a_{15}}
para
a
3
{\displaystyle a_{3}}
a
n
−
p
=
a
m
+
(
n
−
(
m
+
p
)
)
⋅
r
{\displaystyle a_{n-p}=a_{m}+(n-(m+p))\cdot r}
a
n
−
p
=
a
m
+
(
n
−
m
−
p
)
⋅
r
{\displaystyle a_{n-p}=a_{m}+(n-m-p)\cdot r}
a
n
+
p
=
a
m
+
(
n
−
(
m
−
p
)
)
⋅
r
{\displaystyle a_{n+p}=a_{m}+(n-(m-p))\cdot r}
a
n
+
p
=
a
m
+
(
n
−
m
+
p
)
⋅
r
{\displaystyle a_{n+p}=a_{m}+(n-m+p)\cdot r}
a
n
=
a
1
+
(
n
−
1
)
⋅
r
{\displaystyle a_{n}=a_{1}+(n-1)\cdot r}
editar
Podemos escrever uma Progressão Aritmética, tomando como referência o
1
{\displaystyle 1}
º índice:
(
…
,
a
1
−
3
r
⏟
a
−
2
,
a
1
−
2
r
⏟
a
−
1
,
a
1
−
r
⏟
a
0
,
a
1
,
a
1
+
r
⏟
a
2
,
a
1
+
2
r
⏟
a
3
,
a
1
+
3
r
⏟
a
4
,
…
,
a
1
+
(
n
−
3
)
⋅
r
⏟
a
n
−
2
,
a
1
+
(
n
−
2
)
⋅
r
⏟
a
n
−
1
,
a
1
+
(
n
−
1
)
⋅
r
⏟
a
n
,
a
1
+
(
n
−
0
)
⋅
r
⏟
a
n
+
1
,
a
1
+
(
n
+
1
)
⋅
r
⏟
a
n
+
2
,
…
)
{\displaystyle \qquad (\ldots ,\quad \underbrace {a_{1}-3r} _{a_{-2}},\quad \underbrace {a_{1}-2r} _{a_{-1}},\quad \underbrace {a_{1}-r} _{a_{0}},\quad a_{1},\quad \underbrace {a_{1}+r} _{a_{2}},\quad \underbrace {a_{1}+2r} _{a_{3}},\quad \underbrace {a_{1}+3r} _{a_{4}},\quad \ldots ,\quad \underbrace {a_{1}+(n-3)\cdot r} _{a_{n-2}},\quad \underbrace {a_{1}+(n-2)\cdot r} _{a_{n-1}},\quad \underbrace {a_{1}+(n-1)\cdot r} _{a_{n}},\quad \underbrace {a_{1}+(n-0)\cdot r} _{a_{n+1}},\quad \underbrace {a_{1}+(n+1)\cdot r} _{a_{n+2}},\quad \ldots )}
assim obtendo a fórmula para P.A. com referência no
1
{\displaystyle 1}
º índice:
a
n
=
a
1
+
(
n
−
1
)
⋅
r
{\displaystyle a_{n}=a_{1}+(n-1)\cdot r}
ao passar de
a
1
{\displaystyle a_{1}}
para
a
n
{\displaystyle a_{n}}
, avançamos
(
n
−
1
)
{\displaystyle (n-1)}
termos, ou seja, basta somar
(
n
−
1
)
{\displaystyle (n-1)}
vezes a razão ao
1
{\displaystyle 1}
º termo.
a
n
−
p
=
a
n
−
q
−
(
p
−
q
)
r
{\displaystyle a_{n-p}=a_{n-q}-(p-q)r}
editar
a
n
+
p
=
a
n
−
q
+
(
p
+
q
)
r
{\displaystyle a_{n+p}=a_{n-q}+(p+q)r}
editar
Com os índices
n
−
1
{\displaystyle n-1}
e
n
−
2
{\displaystyle n-2}
como referência:
(
…
,
a
n
−
1
−
4
r
⏟
a
n
−
5
,
a
n
−
1
−
3
r
⏟
a
n
−
4
,
a
n
−
1
−
2
r
⏟
a
n
−
3
,
a
n
−
1
−
r
⏟
a
n
−
2
,
a
n
−
1
,
a
n
−
1
+
r
⏟
a
n
,
a
n
−
1
+
2
r
⏟
a
n
+
1
,
a
n
−
1
+
3
r
⏟
a
n
+
2
,
…
)
{\displaystyle (\ldots ,\quad \underbrace {a_{n-1}-4r} _{a_{n-5}},\quad \underbrace {a_{n-1}-3r} _{a_{n-4}},\quad \underbrace {a_{n-1}-2r} _{a_{n-3}},\quad \underbrace {a_{n-1}-r} _{a_{n-2}},\quad a_{n-1},\quad \underbrace {a_{n-1}+r} _{a_{n}},\quad \underbrace {a_{n-1}+2r} _{a_{n+1}},\quad \underbrace {a_{n-1}+3r} _{a_{n+2}},\quad \ldots )}
(
…
,
a
n
−
2
−
4
r
⏟
a
n
−
6
,
a
n
−
2
−
3
r
⏟
a
n
−
5
,
a
n
−
2
−
2
r
⏟
a
n
−
4
,
a
n
−
2
−
r
⏟
a
n
−
3
,
a
n
−
2
,
a
n
−
2
+
r
⏟
a
n
−
1
,
a
n
−
2
+
2
r
⏟
a
n
,
a
n
−
2
+
3
r
⏟
a
n
+
1
,
…
)
{\displaystyle (\ldots ,\quad \underbrace {a_{n-2}-4r} _{a_{n-6}},\quad \underbrace {a_{n-2}-3r} _{a_{n-5}},\quad \underbrace {a_{n-2}-2r} _{a_{n-4}},\quad \underbrace {a_{n-2}-r} _{a_{n-3}},\quad a_{n-2},\quad \underbrace {a_{n-2}+r} _{a_{n-1}},\quad \underbrace {a_{n-2}+2r} _{a_{n}},\quad \underbrace {a_{n-2}+3r} _{a_{n+1}},\quad \ldots )}
deduzindo as fórmulas:
a
n
−
p
=
a
n
−
q
−
(
p
−
q
)
r
{\displaystyle a_{n-p}=a_{n-q}-(p-q)r}
a
n
+
p
=
a
n
−
q
+
(
p
+
q
)
r
{\displaystyle a_{n+p}=a_{n-q}+(p+q)r}
a
n
−
p
=
a
n
+
q
−
(
p
+
q
)
r
{\displaystyle a_{n-p}=a_{n+q}-(p+q)r}
editar
a
n
+
p
=
a
n
+
q
+
(
p
−
q
)
r
{\displaystyle a_{n+p}=a_{n+q}+(p-q)r}
editar
Com os índices
n
+
1
{\displaystyle n+1}
e
n
+
2
{\displaystyle n+2}
como referência:
(
…
,
a
n
+
1
−
4
r
⏟
a
n
−
3
,
a
n
+
1
−
3
r
⏟
a
n
−
2
,
a
n
+
1
−
2
r
⏟
a
n
−
1
,
a
n
+
1
−
r
⏟
a
n
,
a
n
+
1
,
a
n
+
1
+
r
⏟
a
n
+
2
,
a
n
+
1
+
2
r
⏟
a
n
+
3
,
a
n
+
1
+
3
r
⏟
a
n
+
4
,
…
)
{\displaystyle (\ldots ,\quad \underbrace {a_{n+1}-4r} _{a_{n-3}},\quad \underbrace {a_{n+1}-3r} _{a_{n-2}},\quad \underbrace {a_{n+1}-2r} _{a_{n-1}},\quad \underbrace {a_{n+1}-r} _{a_{n}},\quad a_{n+1},\quad \underbrace {a_{n+1}+r} _{a_{n+2}},\quad \underbrace {a_{n+1}+2r} _{a_{n+3}},\quad \underbrace {a_{n+1}+3r} _{a_{n+4}},\quad \ldots )}
(
…
,
a
n
+
2
−
4
r
⏟
a
n
−
2
,
a
n
+
2
−
3
r
⏟
a
n
−
1
,
a
n
+
2
−
2
r
⏟
a
n
,
a
n
+
2
−
r
⏟
a
n
+
1
,
a
n
+
2
,
a
n
+
2
+
r
⏟
a
n
+
3
,
a
n
+
2
+
2
r
⏟
a
n
+
4
,
a
n
+
2
+
3
r
⏟
a
n
+
5
,
…
)
{\displaystyle (\ldots ,\quad \underbrace {a_{n+2}-4r} _{a_{n-2}},\quad \underbrace {a_{n+2}-3r} _{a_{n-1}},\quad \underbrace {a_{n+2}-2r} _{a_{n}},\quad \underbrace {a_{n+2}-r} _{a_{n+1}},\quad a_{n+2},\quad \underbrace {a_{n+2}+r} _{a_{n+3}},\quad \underbrace {a_{n+2}+2r} _{a_{n+4}},\quad \underbrace {a_{n+2}+3r} _{a_{n+5}},\quad \ldots )}
deduzindo as fórmulas:
a
n
−
p
=
a
n
+
q
−
(
p
+
q
)
r
{\displaystyle a_{n-p}=a_{n+q}-(p+q)r}
a
n
+
p
=
a
n
+
q
+
(
p
−
q
)
r
{\displaystyle a_{n+p}=a_{n+q}+(p-q)r}
a
n
−
p
=
a
n
−
p
r
{\displaystyle a_{n-p}=a_{n}-pr}
editar
a
n
+
p
=
a
n
+
p
r
{\displaystyle a_{n+p}=a_{n}+pr}
editar
Com o índice
n
{\displaystyle n}
como referência:
(
…
,
a
n
−
3
r
⏟
a
n
−
3
,
a
n
−
2
r
⏟
a
n
−
2
,
a
n
−
r
⏟
a
n
−
1
,
a
n
,
a
n
+
r
⏟
a
n
+
1
,
a
n
+
2
r
⏟
a
n
+
2
,
a
n
+
3
r
⏟
a
n
+
3
,
…
)
{\displaystyle (\ldots ,\quad \underbrace {a_{n}-3r} _{a_{n-3}},\quad \underbrace {a_{n}-2r} _{a_{n-2}},\quad \underbrace {a_{n}-r} _{a_{n-1}},\quad a_{n},\quad \underbrace {a_{n}+r} _{a_{n+1}},\quad \underbrace {a_{n}+2r} _{a_{n+2}},\quad \underbrace {a_{n}+3r} _{a_{n+3}},\quad \ldots )}
deduzindo as fórmulas:
a
n
−
p
=
a
n
−
p
r
{\displaystyle a_{n-p}=a_{n}-pr}
a
n
+
p
=
a
n
+
p
r
{\displaystyle a_{n+p}=a_{n}+pr}
Demonstração por indução matemática
editar
A fórmula do termo geral pode ser demonstrada por indução matemática:
Ela é válida para o segundo termo pois, por definição, cada termo é igual ao anterior mais uma constante fixa
r
{\displaystyle r}
e portanto
a
2
=
a
1
+
1
⋅
r
{\displaystyle a_{2}=a_{1}+1\cdot r}
Assumindo como hipótese de indução que a fórmula é válida para
n
−
1
{\displaystyle n-1}
, ou seja, que
a
n
−
1
=
a
1
+
(
n
−
2
)
⋅
r
{\displaystyle a_{n-1}=a_{1}+(n-2)\cdot r}
, resulta que o n-ésimo termo é dado por
a
n
=
a
n
−
1
+
r
=
(
a
1
+
(
n
−
2
)
⋅
r
)
+
r
=
a
1
+
(
n
−
2
+
1
)
⋅
r
=
a
1
+
(
n
−
1
)
⋅
r
{\displaystyle \qquad a_{n}=a_{n-1}+r=(a_{1}+(n-2)\cdot r)+r=a_{1}+(n-2+1)\cdot r=a_{1}+(n-1)\cdot r}
Representação por função
editar
Qualquer P.A. pode ser expressa sob a forma de uma função de 1º grau:
f
(
x
)
x
∈
R
a
n
g
e
=
a
x
+
b
{\displaystyle {\underset {x~\in ~Range}{f(x)}}=ax+b}