Question 1201412
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r = common ratio, some nonzero value
a7 = 7th term = 1458
a8 = 8th term = a7*r = 1458r
a9 = 9th term = a8*r = (1458r)*r = 1458r^2
Each time we need a new term, we multiply the previous term by r.


Set the 9th term equal to the stated value 13122 and solve for r.
1458r^2 = 13122
r^2 = 13122/1458
r^2 = 9
r = sqrt(9) or r = -sqrt(9)
r = 3 or r = -3
We have two possibilities for r.
The negative r value will make the geometric sequence terms bounce around from positive to negative, or vice versa.


Consider the case r = 3.
an = a1*(r)^(n-1) 
a7 = a1*(3)^(7-1)
1458 = a1*(3)^6
a1 = 1458/(3^6)
a1 = 2


One possible geometric sequence nth term formula is
an = 2*(3)^(n-1)


Check:
Plug in n = 7
an = 2*(3)^(n-1)
a7 = 2*(3)^(7-1)
a7 = 1458
Repeat for n = 9
an = 2*(3)^(n-1)
a9 = 2*(3)^(9-1)
a9 = 13122
Both values match up, so we've confirmed this formula is correct.


Now consider the case r = -3
an = a1*(r)^(n-1) 
a7 = a1*(-3)^(7-1)
1458 = a1*(-3)^6
But hopefully you can see that (-3)^6 = 3^6 since the exponent is even.
Therefore, we'll land on the same starting a1 value we found earlier (a1 = 2).
We'll arrive at the formula an = 2*(-3)^(n-1)
I'll leave this check section to the student.



We have two possibilities for the nth term formula:
an = 2*(3)^(n-1)
an = 2*(-3)^(n-1)


Let's determine the 2nd term based on those formulas
an = 2*(3)^(n-1)
a2 = 2*(3)^(2-1)
a2 = 6
or
an = 2*(-3)^(n-1)
a2 = 2*(-3)^(2-1)
a2 = -6


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Conclusion:
The 2nd term is 6 if r = 3.
or
The 2nd term is -6 if r = -3.
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