Step-by-step explanation:
Given the Arithmetic sequence
[tex]-7, -3, 1, 5, . . .[/tex]
An arithmetic sequence has a constant difference d and is defined by
[tex]a_n=a_1+\left(n-1\right)d[/tex]
[tex]\mathrm{Compute\:the\:differences\:of\:all\:the\:adjacent\:terms}:\quad \:d=a_{n+1}-a_n[/tex]
[tex]-3-\left(-7\right)=4,\:\quad \:1-\left(-3\right)=4[/tex]
[tex]\mathrm{The\:difference\:between\:all\:of\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}[/tex]
[tex]d=4[/tex]
[tex]\mathrm{The\:first\:element\:of\:the\:sequence\:is}[/tex]
[tex]a_1=-7[/tex]
as
[tex]a_n=a_1+\left(n-1\right)d[/tex]
[tex]\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:[/tex]
[tex]a_n=4\left(n-1\right)-7[/tex] ∵ [tex]d=4[/tex]
[tex]\mathrm{Arithmetic\:sequence\:sum\:formula:}[/tex]
[tex]n\left(a_1+\frac{d\left(n-1\right)}{2}\right)[/tex]
[tex]\mathrm{Plug\:in\:the\:values:}[/tex]
[tex]n=30,\:\space a_1=-7,\:\spaced=4[/tex]
[tex]=30\left(-7+\frac{4\left(30-1\right)}{2}\right)[/tex]
[tex]=30\left(58-7\right)[/tex] ∵ [tex]\frac{4\left(30-1\right)}{2}=58[/tex]
[tex]=30\cdot \:51[/tex]
[tex]=1530[/tex] ∵ [tex]\mathrm{Multiply\:the\:numbers:}\:30\cdot \:51=1530[/tex]
Therefore, the indicated nth partial sum of the arithmetic sequence is 1530.
ANOTHER METHOD
as
[tex]a_n=4\left(n-1\right)-7[/tex]
n = 30
[tex]\sum _{n=1}^{30}\:4\left(n-1\right)-7[/tex]
[tex]=\sum _{n=1}^{30}4n-11[/tex]
[tex]\mathrm{Apply\:the\:Sum\:Rule}:\quad \sum a_n+b_n=\sum a_n+\sum b_n[/tex]
[tex]=\sum _{n=1}^{30}4n-\sum _{n=1}^{30}11[/tex]
as
[tex]\sum _{n=1}^{30}4n=1860[/tex]
and
[tex]\sum _{n=1}^{30}11=330[/tex]
so
[tex]=1860-330[/tex]
[tex]=1530[/tex]
