Respuesta :
Answer:
Explicit formula for
1) [tex]a_n=17-3(n-1)[/tex]
2) [tex]a_n=(\frac{1}{2} )^{n-1}\cdot 20[/tex]
Step-by-step explanation:
W have to find the first few terms of the given sequence and then find an explicit equation
Explicit formula = f(n) + d(n-1) , where,
f(n) is first term,
d is common difference,
n-1 is one term less than the term number.
1)
Given : [tex]a_1=17\\\\a_{n+1}=a_n-3[/tex]
[tex]\text{We put n =1, we get,}\\\\a_2=a_1-3=17-3=11=17-3(2-1)\\\\\\\text{We put n =2, we get,}\\\\\\a_3=a_2-3=11-3=8=17-3-3=17-3\cdot 2=17-3(3-1)\\\\\text{We put n =3, we get,}\\\\\\a_4=a_3-3=8-3=5=17-3-3-3=17-3\cdot 3=17-3(4-1)\\[/tex]
Thus, we obtained an explicit formula,
[tex]a_n=17-3(n-1)[/tex]
2)
Given : [tex]a_1=20\\\\a_{n+1}=\frac{1}{2} \cdot a_n[/tex]
[tex]\text{We put n =1, we get,}\\\\a_2=a_1-3=17-3=11=17-3(2-1)\\\\\\\text{We put n =2, we get,}\\\\\\a_3=a_2-3=11-3=8=17-3-3=17-3\cdot 2=17-3(3-1)\\\\\text{We put n =3, we get,}\\\\\\a_4=a_3-3=8-3=5=17-3-3-3=17-3\cdot 3=17-3(4-1)\\[/tex]
Thus, we obtained an explicit formula,
[tex]a_n=17-3(n-1)[/tex]
[tex]\text{We put n =1, we get,}\\\\\\a_2=\frac{1}{2} \cdot a_1=\frac{1}{2} \cdot 20=10=(\frac{1}{2} )^{2-1}\cdot 20\\\\\\\text{We put n =3, we get,}\\\\\\a_4=\frac{1}{2} \cdot a_3=\frac{1}{2} \cdot 5=\frac{5}{2}=(\frac{1}{2} )^{4-1}\cdot 20\\\\\\[/tex]
Thus, we obtained an explicit formula,
[tex]a_n=(\frac{1}{2} )^{n-1}\cdot 20[/tex]
Answer:
1. Terms are 17, 14, 11, 8, 5,...... and explicit equation is [tex]a_{n}=17-3(n-1)[/tex].
2. Terms are 20, 10, 5, 2.5, 1.25,...... and explicit equation is [tex]a_{n}=20\times (\frac{1}{2})^{n-1}[/tex].
Step-by-step explanation:
Ques 1: We are given the recursive formula for the sequence as,
[tex]a_{n+1}=a_{n}-3[/tex], where [tex]a_{1}=17[/tex].
So, substituting the values of 'n' from {1,2,3,.....}, we get,
[tex]a_{2}=a_{1+1}=a_{1}-3=17-3=14[/tex]
[tex]a_{3}=a_{2+1}=a_{2}-3=14-3=11[/tex]
[tex]a_{4}=a_{3+1}=a_{3}-3=11-3=8[/tex]
[tex]a_{5}=a_{4+1}=a_{4}-3=8-3=5[/tex]
Thus, the sequence is given by 17, 14, 11, 8, 5,......
As, the explicit equation of an arithmetic sequence is of the form, [tex]a_{n}=a_{1}+d(n-1)[/tex], where [tex]a_{1}[/tex] is the first term and 'd' is the common difference.
As, the common difference, d = 14 - 17 = -3
Thus, we get,
The given sequence has the explicit equation, [tex]a_{n}=17-3(n-1)[/tex].
Ques 2: We are given the recursive formula for the sequence as,
[tex]a_{n+1}=\frac{a_{n}}{2}[/tex], where [tex]a_{1}=20[/tex].
So, substituting the values of 'n' from {1,2,3,.....}, we get,
[tex]a_{2}=a_{1+1}=\frac{a_{1}}{2}=\frac{20}{2}=10[/tex]
[tex]a_{3}=a_{2+1}=\frac{a_{2}}{2}=\frac{10}{2}=5[/tex]
[tex]a_{4}=a_{3+1}=\frac{a_{3}}{2}=\frac{5}{2}=2.5[/tex]
[tex]a_{5}=a_{4+1}=\frac{a_{4}}{2}=\frac{2.5}{2}=1.25[/tex]
Thus, the sequence is given by 20, 10, 5, 2.5, 1.25,......
As, the explicit equation of a geometric sequence is of the form, [tex]a_{n}=a_{1}\times r^(n-1)[/tex], where [tex]a_{1}[/tex] is the first term and 'r' is the common ratio.
As, the common ratio, [tex]r=\frac{10}{20}=\frac{1}{2}[/tex]
Thus, we get,
The given sequence has the explicit equation, [tex]a_{n}=20\times (\frac{1}{2})^{n-1}[/tex].
