Find the explicit formula of the arithmetic sequences defined as follows. The first term of each sequence is a_1.
{\begin{cases} a_1=4 \cr \cr\forall n \in \mathbb{N}^*,\ a_{n+1}=a_n-2 \end{cases}\\}
We have:
a_{n+1}=a_n-2
a_{n+1}-a_n=-2
The common difference of the sequence is -2.
The following is the explicit formula of an arithmetic sequence whose first term is a_1 and common difference is d :
a_n= a_1+d\left(n-1\right)
Thus, we can write:
a_n = 4+ \left(-2\right)\left(n-1\right) \\= 4-2n+2\\=-2n+6
For n≥1, a_n=-2n+6
\begin{cases} a_3=15 \cr \cr \forall n \in \mathbb{N}^*,\ a_{n+1}=a_n+4 \end{cases}\\
We have:
a_{n+1}=a_n+4
a_{n+1}-a_n=4
The common difference of the sequence is 4.
The following is the explicit formula of an arithmetic sequence whose first term is a_1 and common difference is d :
a_n= a_1+d\left(n-1\right).
To find the explicit formula, find a_1. For n=3 we have:
a_3= a_1+d\left(3-1\right) \\\Rightarrow 15= a_1 + 4 \times 2\\\Rightarrow a_1=7
Thus, we can write:
a_n = 7+ 4\left(n-1\right) \\= 7+4n-4\\=4n+3
For n≥1, a_n=4n+3
\begin{cases} a_1=-1 \cr \cr \forall n \geq2,\ a_{n+1}=a_{n-1}+6 \end{cases}\\
We have:
a_{n+1}=a_{n-1}+6
a_{n+1}-a_{n-1}=6\\\\\left(a_{n+1}-a_n\right)+\left(a_n-a_{n-1}\right)=6\\d+d = 6\\d=3
a_{n+1}-a_{n-1}=6
\left(a_{n+1}-a_n\right)+\left(a_n-a_{n-1}\right)=6
d+d = 6
d=3
The common difference of the sequence is 3.
The following is the explicit formula of an arithmetic sequence whose first term is a_1 and common difference is d :
a_n= a_1+d\left(n-1\right)
Thus, we can write:
a_n = -1+ 3\left(n-1\right) \\= -1+3n-3\\=3n-4
For n≥1, a_n=3n-4
\begin{cases} a_1=-3 \cr \cr\forall n \in \mathbb{N}^*,\ a_{n+1}=a_n+5 \end{cases}\\
We have:
a_{n+1}=a_n+5
a_{n+1}-a_n=5
The common difference of the sequence is 5.
The following is the explicit formula of an arithmetic sequence whose first term is a_1 and common difference is d :
a_n= a_1+d\left(n-1\right)
Thus, we can write:
a_n = -3+ 5\left(n-1\right) \\= -3+5n-5\\=5n-8
For n≥1, a_n=5n-8
\begin{cases} a_1=0 \cr \cr\forall n \in \mathbb{N}^*,\ a_{n+2}=a_n-10 \end{cases}\\
We have:
a_{n+2}=a_{n}-10
And:
a_{n+2}-a_{n}=\left(a_{n+2}-a_{n+1}\right)+\left(a_{n+1}-a_{n-1}\right)
a_{n+2}-a_{n}=d+d = 2d
Therefore:
2d=-10
d=-5
The common difference of the sequence is -5.
The following is the explicit formula of an arithmetic sequence whose first term is a_1 and common difference is d :
a_n= a_1+d\left(n-1\right)
Thus, we can write:
a_n = 0-5\left(n-1\right) \\= -5n+5
For n≥1, a_n=-5n+5
\begin{cases} a_5=0 \cr \cr\forall n \in \mathbb{N}^*,\ a_{n+1}=a_n+5 \end{cases}\\
We have:
a_{n+1}=a_n+5
a_{n+1}-a_n=5
The common difference of the sequence is 5.
The following is the explicit formula of an arithmetic sequence whose first term is a_1 and common difference is d :
a_n= a_1+d\left(n-1\right)
To find the explicit formula, find a_1. For n=5 we have:
a_5 = a_1+ d\left(5-1\right)
0= a_1+5\left(4\right)
0=a_1+20
a_1=-20
Thus, we can write:
a_n = -20+ 5\left(n-1\right) \\= -20+5n-5\\\\=5n-25
For n≥1, a_n=5n-25
\begin{cases} a_1=-4 \cr \cr\forall n\geq2,\ a_{n+5}=a_{n-1}+18 \end{cases}\\
For two positive integers m and n, use the following formula to find the common difference of the sequence:
d=\dfrac{a_n-a_m}{n-m}
In this question we have:
a_{n+5}=a_{n-1}+18
a_{n+6}-a_{n-1}=18
Thus, we can write:
d=\dfrac{a_{n+5}-a_{n-1}}{\left(n+5\right)-\left(n-1\right)}
d=\dfrac{18}{\left(n+5-n+1\right)}\\
d=\dfrac{18}{6}=3
The common difference of the sequence is 3.
The following is the explicit formula of an arithmetic sequence whose first term is a_1 and common difference is d :
a_n= a_1+d\left(n-1\right)
Thus, we can write:
a_n = -4+ 3\left(n-1\right) \\=- 4+3n-3\\=3n-7
For n≥1, a_n=3n-7