Question

Find an equation for the nth term of a geometric sequence where the second and fifth terms are -21 and 567, respectively.

Answer 1

Answer:

$a_n=7 \cdot (-3)^{n-1}$

Step-by-step explanation:

The explicit form for a geometric sequence is $a_n=a_1 \cdot r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio.

We have the following given:

$a_2=-21$

$a_5=567$.

We also know that $a_2=a_1 \cdot r$ while $a_5=a_1 \cdot r_4$.

So if we do 5th term divided by second term we get:

$\frac{a_1 \cdot r_4}{a_1 \cdot r}=\frac{567}{-21}$

Simplifying both sides:

$r^3=-27$

Cube root both sides:

$r=-3$

The common ratio, r, is -3.

Now we need to find the first term.

That shouldn't be too hard here since we know the second term which is -21.

We know that first term times the common ratio will give us the second term.

So we are solving the equation:

$a_1 \cdot r=a_2$.

$a_1 \cdot (-3)=-21$

Dividing both sides by -3 gives us $a_1=7$.

So the equation is in it's explicit form is:

$a_n=7 \cdot (-3)^{n-1}$

Check it!

Plugging in 2 should gives us a result of -21.

$a_2=7 \cdot (-3)^{2-1}$

$a_2=7 \cdot (-3)^1$

$a_2=7 \cdot (-3)$

$a_2=-21$

That checks out!

Plugging in 5 should give us a result of 567.

$a_5=7 \cdot (-3)^{5-1}$

$a_5=7 \cdot (-3)^4$

$a_5=7 \cdot 81$

$a_5=567$

The checks out!

Our equation works!