Question

Find the explicit formula for the general nth term of the geometric sequence described below.


a4= -32 and a7= -256

an=?

Answer 1

Answer:

$a_n=-4(2)^{n-1}$

Step-by-step explanation:

Let the geometric sequence be:

First term: $a_1=a$

Second term: $a_2=ar$

Third term: $a_3=ar^2$

Fourth term:  $a^4=ar^3$

Follow the pattern for the terms between the 4th and 7th: ..........

Seventh term: $a_7=ar^6$

Follow the pattern for the terms between the 7th and the nth: .........

$a_n=ar^{n-1}$

So we need to find $a$ which the value of the first term.

We need to find $r$ which is the common ratio here.

So if we take the 7th term and divide the 4th term we get the following equation:

$\frac{a_7}{a^4}=\frac{-256}{-32}$

$\frac{ar^6}{ar^3}=\frac{-256}{-32}$

Now the first thing I notice is on the left hand side. I can cancel out some common factors over the numerator and denominator:

$r^{6-3}=8$

$r^{3}=8$

Now I see to find the common ratio, $r$, we can just take the cube root of both sides:

$r=\sqrt[3]{8}$

$r=2$ is true since 2(2)(2)=8.

So after finding $r$ has value 2, we need to find $a$ which is the first term of the sequence.

Let's use one of our equations with a given value for it:

$a_4=-32$

$ar^3=-32$

$ar^3=-32$ with $r=2$:

$a(2)^3=-32$

$a(8)=-32$

Divide both sides by 8:

$a=\frac{-32}{8}$

$a=-4$

So the first term,$a$, has value -4.

So the explicit form of this geometric sequence is:

$a_n=-4(2)^{n-1}$

Let's verify:

What happens when $n=4$?

$a_4=-4(2)^{4-1}$

$a_4=-4(2)^3$

$a_4=-4(8)$

$a_4=-32$

That look's good.

Let's check the other given condition.

$n=7$?

$a_7=-4(2)^{7-1}$

$a_7=-4(2)^{6}$

$a_7=-4(64)$

$a_7=-256$

Since both of the conditions are satisfied, we have done our job and it is confirmed.

Answer 2

Answer:

An=-4(2)^n-1

Step-by-step explanation: