Respuesta :
This is a geometric sequence, with the first term: 1/64
The common ratio is found through the ratio of the first two terms and the second and third term.
[tex]\frac{\frac{1}{32}}{\frac{1}{64}} = \frac{\frac{1}{16}}{\frac{1}{32}}[/tex]
[tex]\frac{1}{32} \cdot 64 = \frac{1}{16} \cdot 32[/tex]
Since there is a common difference of 2, we can generalise this sequence to find the nth term:
[tex]T_n = ar^{n-1}[/tex]
a is the first term, r is the common ratio, and n is the nth term in the sequence.
[tex]T_n = \frac{1}{64} \cdot 2^{n-1}[/tex]
Substituting n = 7, we get:
[tex]T_7 = \frac{1}{64} \cdot 2^{7 - 1}[/tex]
[tex]T_7 = \frac{1}{64} \cdot 64[/tex]
[tex]T_7 = \frac{64}{64} = 1[/tex]
Thus, the next term in the sequence is 1.
The common ratio is found through the ratio of the first two terms and the second and third term.
[tex]\frac{\frac{1}{32}}{\frac{1}{64}} = \frac{\frac{1}{16}}{\frac{1}{32}}[/tex]
[tex]\frac{1}{32} \cdot 64 = \frac{1}{16} \cdot 32[/tex]
Since there is a common difference of 2, we can generalise this sequence to find the nth term:
[tex]T_n = ar^{n-1}[/tex]
a is the first term, r is the common ratio, and n is the nth term in the sequence.
[tex]T_n = \frac{1}{64} \cdot 2^{n-1}[/tex]
Substituting n = 7, we get:
[tex]T_7 = \frac{1}{64} \cdot 2^{7 - 1}[/tex]
[tex]T_7 = \frac{1}{64} \cdot 64[/tex]
[tex]T_7 = \frac{64}{64} = 1[/tex]
Thus, the next term in the sequence is 1.
