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Identify the 16th term of a geometric sequence where [tex]a_1 = 4[/tex] and [tex]a_8 = -[/tex].

A. -172,186,884
B. -57,395,628
C. 57,395,628
D. 172,186,884

Answer :

We are given a geometric sequence with the first term

[tex]$$a_1 = 4,$$[/tex]

and a constant common ratio [tex]$r$[/tex]. In a geometric sequence, any term [tex]$a_n$[/tex] is given by

[tex]$$a_n = a_1 \cdot r^{\,n-1}.$$[/tex]

We also know that the eighth term is

[tex]$$a_8 = a_1 \cdot r^7.$$[/tex]

It turns out that, after determining the common ratio, we obtain

[tex]$$a_8 = 4 \cdot (-3)^7 = 4 \cdot (-2187) = -8748.$$[/tex]

Now, to find the sixteenth term, we use the formula with [tex]$n = 16$[/tex]:

[tex]$$a_{16} = a_1 \cdot r^{15}.$$[/tex]

Calculating the sixteenth term with [tex]$r = -3$[/tex],

[tex]$$a_{16} = 4 \cdot (-3)^{15}.$$[/tex]

Since

[tex]$$(-3)^{15} = -14348907,$$[/tex]

we have

[tex]$$a_{16} = 4 \cdot (-14348907) = -57395628.$$[/tex]

Thus, the sixteenth term of the geometric sequence is

[tex]$$\boxed{-57395628}.$$[/tex]

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