Answer:
The ratio of all the adjacent terms is the same and equal to
[tex]r=-4[/tex]
The next three terms after the last one will be:
Step-by-step explanation:
Given the sequence
-4, 16, -64, 256, …
Finding the common ratio
An arithmetic sequence has a constant ratio 'r' and is defined by
[tex]a_n=a_1\cdot r^{n-1}[/tex]
computing the ratios of all the adjacent terms
[tex]\frac{16}{-4}=-4,\:\quad \frac{-64}{16}=-4,\:\quad \frac{256}{-64}=-4[/tex]
The ratio of all the adjacent terms is the same and equal to
[tex]r=-4[/tex]
Finding the next three terms
Given the sequence
-4, 16, -64, 256, …
here
[tex]a_1=-4[/tex]
[tex]r=-4[/tex]
substituting [tex]a_1=-4[/tex] and [tex]r=-4[/tex] in the nth term
[tex]a_n=a_1\cdot r^{n-1}[/tex]
[tex]a_n=-4\left(-4\right)^{n-1}[/tex]
substituting n = 5 to determine the 5th term
[tex]a_5=-4\left(-4\right)^{5-1}[/tex]
[tex]a_5=-4^4\cdot \:4[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}[/tex]
[tex]a_5=-4^{1+4}[/tex]
[tex]a_5=-4^5[/tex]
[tex]a_5=-1024[/tex]
substituting n = 6 to determine the 6th term
[tex]a_6=-4\left(-4\right)^{6-1}[/tex]
[tex]a_6=-4\left(-4^5\right)[/tex]
[tex]\mathrm{Apply\:rule}\:-\left(-a\right)=a[/tex]
[tex]a_6=4\cdot \:4^5[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}[/tex]
[tex]a_6=4^{1+5}[/tex]
[tex]a_6=4^6[/tex]
[tex]a_6=4096[/tex]
substituting n = 7 to determine the 6th term
[tex]a_7=-4\left(-4\right)^{7-1}[/tex]
[tex]a_7=-4^6\cdot \:4[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}[/tex]
[tex]a_7=-4^{1+6}[/tex]
[tex]a_7=-4^7[/tex]
[tex]a_7=-16384[/tex]
Therefore, the next three terms after the last one will be:
