Respuesta :
An arithmetic sequence has the recursive structure
[tex]a_n=a_{n-1}+d[/tex]
and can be solved explicitly for the [tex]n[/tex] term, [tex]a_n[/tex], in terms of the first term of the sequence [tex]a_1[/tex].
[tex]a_n=a_{n-1}+d[/tex]
[tex]\implies~a_n=a_{n-2}+2d[/tex]
[tex]\implies~a_n=a_{n-3}+3d[/tex]
[tex]\implies\cdots\implies a_n=a_1+(n-1)d[/tex]
You know the 100th term of the sequence is [tex]a_{100}=1000[/tex] and that the common difference between terms is [tex]d=5[/tex], which means you have enough information to find the first term:
[tex]a_{100}=a_1+(100-1)\times5\implies1000=a_1+99\times5\implies a_1=505[/tex]
[tex]a_n=a_{n-1}+d[/tex]
and can be solved explicitly for the [tex]n[/tex] term, [tex]a_n[/tex], in terms of the first term of the sequence [tex]a_1[/tex].
[tex]a_n=a_{n-1}+d[/tex]
[tex]\implies~a_n=a_{n-2}+2d[/tex]
[tex]\implies~a_n=a_{n-3}+3d[/tex]
[tex]\implies\cdots\implies a_n=a_1+(n-1)d[/tex]
You know the 100th term of the sequence is [tex]a_{100}=1000[/tex] and that the common difference between terms is [tex]d=5[/tex], which means you have enough information to find the first term:
[tex]a_{100}=a_1+(100-1)\times5\implies1000=a_1+99\times5\implies a_1=505[/tex]
The nth term of an arithmetic sequence = a + (n - 1)d
a = 1st term = ? n =nth term = 100. d = common difference = 5
a₁₀₀ = a + (100 - 1)d
a₁₀₀ = a + 99d = 1000
a + 99d = 1000
a + 99*5 = 1000
a + 495 = 1000
a = 1000 - 495
a = 505
So the first term a, is the same as a₁ = 505
I hope this helped.
a = 1st term = ? n =nth term = 100. d = common difference = 5
a₁₀₀ = a + (100 - 1)d
a₁₀₀ = a + 99d = 1000
a + 99d = 1000
a + 99*5 = 1000
a + 495 = 1000
a = 1000 - 495
a = 505
So the first term a, is the same as a₁ = 505
I hope this helped.