What is the 6th term of the geometric sequence where a1 = –4,096 and a4 = 64?

1 Answer
Nov 6, 2015



a*n : so for term a1, n=1

We know that the signs of the terms alternate from negative on odd values of n to positive on even values of n. We need to account for this and using an exponential will be an easy solution. If we use (-1)^n, the whenever n is odd (such as a1) this will return a negative. If n is even (such as a4), then we get a positive.

(-1)^1 = -1
(-1)^4 = 1

Now we also see as n increases, our terms decrease in value. When you divide a number by an increasingly large number (n in this case), the over all value will decrease. So we know that our geometric sequence will be structured around a fraction with n in the denominator.

Now we play with the numbers.

If a1=-4096 and n=1 and we need it to be in fraction form then:

a*n= (-1)^n [4095/n] works

If a4=64 and n=4, let's try this in our equation from above:

a*4= (-1)^4 [4096/4] = 1024

This is NOT 64, but we see that 64 goes into 1024 evenly.

1024/64= 16, and 16=4*4

To reflect this in our sequence, we need to add this in the denominator so:

a4= (-1)^4 [4096/(44*4)] =64

So: a*n= (-1)^n [4096/n^3]

Plugging n=1 and n=4 into this new equation, we see it satifies our conditions for both a1 and a4.

Now we solve for when n=6 and we get 18.96