Question

Find the common ratio and the three terms in the sequence after the last one given. -4, 16, -64, 256, …

Answer 1

Answer:

The ratio of all the adjacent terms is the same and equal to

$r=-4$

The next three terms after the last one will be:

• $a_5=-1024$

• $a_6=4096$

• $a_7=-16384$

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  

$a_n=a_1\cdot r^{n-1}$

computing the ratios of all the adjacent terms

$\frac{16}{-4}=-4,\:\quad \frac{-64}{16}=-4,\:\quad \frac{256}{-64}=-4$

The ratio of all the adjacent terms is the same and equal to

$r=-4$

Finding the next three terms

Given the sequence

-4, 16, -64, 256, …

here

$a_1=-4$

$r=-4$

substituting $a_1=-4$ and $r=-4$ in the nth term

$a_n=a_1\cdot r^{n-1}$

$a_n=-4\left(-4\right)^{n-1}$

substituting n = 5 to determine the 5th term

$a_5=-4\left(-4\right)^{5-1}$

$a_5=-4^4\cdot \:4$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}$

$a_5=-4^{1+4}$

$a_5=-4^5$

$a_5=-1024$

substituting n = 6 to determine the 6th term

$a_6=-4\left(-4\right)^{6-1}$

$a_6=-4\left(-4^5\right)$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a$

$a_6=4\cdot \:4^5$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}$

$a_6=4^{1+5}$

$a_6=4^6$

$a_6=4096$

substituting n = 7 to determine the 6th term

$a_7=-4\left(-4\right)^{7-1}$

$a_7=-4^6\cdot \:4$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}$

$a_7=-4^{1+6}$

$a_7=-4^7$

$a_7=-16384$

Therefore, the next three terms after the last one will be:

• $a_5=-1024$

• $a_6=4096$

• $a_7=-16384$