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