Answer:
[tex]\textsf{Formula 1:}\quad a_n=4(-3)^{n-1}\quad a_7=2916[/tex]
[tex]\textsf{Formula 2:}\quad a_n=4(2)^{n-1}\quad a_7=256[/tex]
Step-by-step explanation:
The general form of the nth term of a geometric sequence is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{General form of the $n$th term of an geometric sequence}}\\\\a_n=ar^{n-1}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a_n$ is the $n$th term.}\\\phantom{ww}\bullet \;\textsf{$a$ is the first term.}\\\phantom{ww}\bullet\;\textsf{$r$ is the common ratio.}\\\phantom{ww}\bullet\;\textsf{$n$ is the position of the term.}\end{array}}[/tex]
Given that the first term in the sequence is 4, then:
[tex]a_n=4r^{n-1}[/tex]
Given that the sum of the first three terms is 28, then:
[tex]a_1+a_2+a_3=28\\\\\\4+4r^{2-1}+4r^{3-1}=28\\\\\\4+4r^{1}+4r^{2}=28\\\\\\4+4r+4r^{2}=28\\\\\\4(1+r+r^2)=4(7)\\\\\\1+r+r^2=7\\\\\\r^2+r-6=0[/tex]
Solve for r:
[tex]r^2+3r-2r-6=0\\\\\\r(r+3)-2(r+3)=0\\\\\\(r+3)(r-2)=0\\\\\\r=-3,\;\;r=2[/tex]
So, we have two possible common ratios, r = -3 and r = 2.
If r = -3, then the sequence alternates signs:
[tex]a_1=4\\\\a_2=4(-3)^1=-12\\\\a_3=4(-3)^2=36\\\\a_4=4(-3)^3=-108[/tex]
If r = 2, then:
[tex]a_1=4\\\\a_2=4(2)^1=8\\\\a_3=4(2)^2=16\\\\a_4=4(2)^3=32[/tex]
The two possible explicit formulas for the sequence given the two found common ratios are:
[tex]a_n=4(-3)^{n-1}\\\\a_n=4(2)^{n-1}[/tex]
To find the seventh term, we can substitute n = 7 into the formulas:
[tex]a_7=4(-3)^{7-1}\\\\a_7=4(-3)^6\\\\a_7=4(729)\\\\a_7=2916\\\\\\\\a_7=4(2)^{7-1}\\\\a_7=4(2)^6\\\\a_7=4(64)\\\\a_7=256[/tex]
Therefore, the seventh terms are 2,916 and 256.
