Question
The seventh term of the gp 16/9, -8/3, 4........is?
Answers
GPT-4o mini
To find the seventh term of the geometric progression (GP) where the first few terms are given as \( \frac{16}{9}, -\frac{8}{3}, 4 \), we first need to determine the common ratio \( r \) of the GP.
We can find the common ratio by dividing the second term by the first term:
\[
r = \frac{\text{second term}}{\text{first term}} = \frac{-\frac{8}{3}}{\frac{16}{9}} = -\frac{8}{3} \times \frac{9}{16} = -\frac{72}{48} = -\frac{3}{2}
\]
Now let's verify this common ratio by checking the ratio of the third term to the second term:
\[
\text{third term} = 4
\]
Calculating the ratio:
\[
r = \frac{\text{third term}}{\text{second term}} = \frac{4}{-\frac{8}{3}} = 4 \times -\frac{3}{8} = -\frac{12}{8} = -\frac{3}{2}
\]
This confirms that the common ratio \( r \) is indeed \( -\frac{3}{2} \).
Now, for a geometric progression, the \(n\)th term can be calculated using the formula:
\[
a_n = a_1 \cdot r^{n-1}
\]
where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.
Here, the first term \( a_1 = \frac{16}{9} \), \( r = -\frac{3}{2} \), and we want the seventh term (\( n = 7 \)):
\[
a_7 = a_1 \cdot r^{7-1} = \frac{16}{9} \cdot r^{6} = \frac{16}{9} \cdot \left(-\frac{3}{2}\right)^{6}
\]
Calculating \( \left(-\frac{3}{2}\right)^6 \):
\[
\left(-\frac{3}{2}\right)^6 = \left(\frac{3}{2}\right)^6 = \frac{729}{64}
\]
Now substituting this value back into the equation for the seventh term:
\[
a_7 = \frac{16}{9} \cdot \frac{729}{64}
\]
Calculating this product:
\[
a_7 = \frac{16 \cdot 729}{9 \cdot 64}
\]
Now simplifying:
1. Simplify \( \frac{16}{64} = \frac{1}{4} \).
2. Then \( a_7 = \frac{1 \cdot 729}{4 \cdot 9} = \frac{729}{36} \).
Finally, simplify \( \frac{729}{36} \):
\[
\frac{729 \div 9}{36 \div 9} = \frac{81}{4}
\]
Thus, the seventh term of the GP is
\[
\boxed{\frac{81}{4}}.
\]
We can find the common ratio by dividing the second term by the first term:
\[
r = \frac{\text{second term}}{\text{first term}} = \frac{-\frac{8}{3}}{\frac{16}{9}} = -\frac{8}{3} \times \frac{9}{16} = -\frac{72}{48} = -\frac{3}{2}
\]
Now let's verify this common ratio by checking the ratio of the third term to the second term:
\[
\text{third term} = 4
\]
Calculating the ratio:
\[
r = \frac{\text{third term}}{\text{second term}} = \frac{4}{-\frac{8}{3}} = 4 \times -\frac{3}{8} = -\frac{12}{8} = -\frac{3}{2}
\]
This confirms that the common ratio \( r \) is indeed \( -\frac{3}{2} \).
Now, for a geometric progression, the \(n\)th term can be calculated using the formula:
\[
a_n = a_1 \cdot r^{n-1}
\]
where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.
Here, the first term \( a_1 = \frac{16}{9} \), \( r = -\frac{3}{2} \), and we want the seventh term (\( n = 7 \)):
\[
a_7 = a_1 \cdot r^{7-1} = \frac{16}{9} \cdot r^{6} = \frac{16}{9} \cdot \left(-\frac{3}{2}\right)^{6}
\]
Calculating \( \left(-\frac{3}{2}\right)^6 \):
\[
\left(-\frac{3}{2}\right)^6 = \left(\frac{3}{2}\right)^6 = \frac{729}{64}
\]
Now substituting this value back into the equation for the seventh term:
\[
a_7 = \frac{16}{9} \cdot \frac{729}{64}
\]
Calculating this product:
\[
a_7 = \frac{16 \cdot 729}{9 \cdot 64}
\]
Now simplifying:
1. Simplify \( \frac{16}{64} = \frac{1}{4} \).
2. Then \( a_7 = \frac{1 \cdot 729}{4 \cdot 9} = \frac{729}{36} \).
Finally, simplify \( \frac{729}{36} \):
\[
\frac{729 \div 9}{36 \div 9} = \frac{81}{4}
\]
Thus, the seventh term of the GP is
\[
\boxed{\frac{81}{4}}.
\]