Answer :
To identify the 16th term of a geometric sequence where the first term [tex]\(a_1 = 4\)[/tex] and the 8th term [tex]\(a_8 = -8,748\)[/tex], follow these steps:
1. Understand the formula for the nth term of a geometric sequence:
The formula is given by:
[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.
2. Use the formula to find the common ratio [tex]\(r\)[/tex]:
For the 8th term:
[tex]\[
a_8 = a_1 \times r^{(8-1)} = 4 \times r^7 = -8,748
\][/tex]
To find [tex]\(r\)[/tex], solve the equation:
[tex]\[
r^7 = \frac{-8,748}{4}
\][/tex]
[tex]\[
r^7 = -2,187
\][/tex]
So, [tex]\(r\)[/tex] is the 7th root of [tex]\(-2,187\)[/tex].
3. Calculate the 16th term using the found common ratio [tex]\(r\)[/tex]:
Using the formula for the nth term again:
[tex]\[
a_{16} = a_1 \times r^{(16-1)} = 4 \times r^{15}
\][/tex]
4. Substitute the common ratio into the equation for the 16th term:
After solving for the common ratio and substituting back, we find:
[tex]\[
a_{16} \approx -57,395,628
\][/tex]
So, the 16th term of the sequence is [tex]\(-57,395,628\)[/tex].
1. Understand the formula for the nth term of a geometric sequence:
The formula is given by:
[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.
2. Use the formula to find the common ratio [tex]\(r\)[/tex]:
For the 8th term:
[tex]\[
a_8 = a_1 \times r^{(8-1)} = 4 \times r^7 = -8,748
\][/tex]
To find [tex]\(r\)[/tex], solve the equation:
[tex]\[
r^7 = \frac{-8,748}{4}
\][/tex]
[tex]\[
r^7 = -2,187
\][/tex]
So, [tex]\(r\)[/tex] is the 7th root of [tex]\(-2,187\)[/tex].
3. Calculate the 16th term using the found common ratio [tex]\(r\)[/tex]:
Using the formula for the nth term again:
[tex]\[
a_{16} = a_1 \times r^{(16-1)} = 4 \times r^{15}
\][/tex]
4. Substitute the common ratio into the equation for the 16th term:
After solving for the common ratio and substituting back, we find:
[tex]\[
a_{16} \approx -57,395,628
\][/tex]
So, the 16th term of the sequence is [tex]\(-57,395,628\)[/tex].
