Question

Write a sequence that has four geometric means between 31 and –23,540,625.

Answer 1

Answer:

$a_n=31(-15)^{n-1}$

31, -465, 6975, -104,625, 1,569,375, -23,540,625

Step-by-step explanation:

The formula for a geometric sequence is:

$a_n=a_1(r)^{n-1}$

The formula for a geometric sequence is:

Where

r is the common ratio

$a_1$ is the first term

$a_n$ is the nth term

In this case

$a_1=31$

$a_6=-23,540,625$

So:

$-23,540,625=31(r)^{6-1}$

Now we solve for r

$-23,540,625=31(r)^{5}$

$-\frac{23,540,625}{31}=r^{5}\\\\r=\sqrt[5]{-\frac{23,540,625}{31}}\\\\r=-15$

Then the four geometric means are

$a_2=31(-15)^{2-1}=-465$

$a_3=31(-15)^{3-1}=6975$

$a_4=31(-15)^{4-1}=-104,625$

$a_5=31(-15)^{5-1}=1,569,375$

31, -465, 6975, -104,625, 1,569,375, -23,540,625

Answer 2

Answer:

31, -465, 6975, -104,625, 1,569,375, -23,540,625

Step-by-step explanation: