College

Identify the 16th term of a geometric sequence where [tex]$a_1 = 4$[/tex] and [tex]$a_8 = -8,748$[/tex].

A. [tex]-172,186,884[/tex]
B. [tex]-57,395,628[/tex]
C. [tex]57,395,628[/tex]
D. [tex]172,186,884[/tex]

Answer :

To identify the 16th term of a geometric sequence, we first need to understand the elements of the sequence. We're given that the first term [tex]\( a_1 \)[/tex] is 4 and the 8th term [tex]\( a_8 \)[/tex] is -8,748 in the sequence.

A geometric sequence is defined by the relationship:
[tex]\[ a_n = a_1 \times r^{n-1} \][/tex]
where [tex]\( a_n \)[/tex] is the nth term, [tex]\( a_1 \)[/tex] is the first term, and [tex]\( r \)[/tex] is the common ratio.

Step 1: Find the common ratio [tex]\( r \)[/tex].

Given:
[tex]\[ a_8 = -8,748 \][/tex]
[tex]\[ a_1 = 4 \][/tex]
We use the formula for the 8th term:
[tex]\[ a_8 = a_1 \times r^{8-1} \][/tex]
[tex]\[ -8,748 = 4 \times r^7 \][/tex]

To find [tex]\( r \)[/tex], rearrange the equation:
[tex]\[ r^7 = \frac{-8,748}{4} = -2,187 \][/tex]
[tex]\[ r = (-2,187)^{1/7} \][/tex]

Step 2: Find the 16th term, [tex]\( a_{16} \)[/tex].

Now, use the formula for the nth term again:
[tex]\[ a_{16} = a_1 \times r^{16-1} \][/tex]
[tex]\[ a_{16} = 4 \times r^{15} \][/tex]

Given the previously calculated [tex]\( r \)[/tex] and knowing the values obtained (treating them as accurate for this problem):
[tex]\[ a_{16} \approx 57,395,628 \][/tex]

Thus, the 16th term of the sequence is approximately [tex]\( 57,395,628 \)[/tex]. Therefore, the answer is:
[tex]\[ 57,395,628 \][/tex]