What is the 7th term of the geometric sequence where a1 = 1,024 and a4 = -16?

1 Answer
May 14, 2016

Answer:

#a_7 = 1024 (1/-4)^(6) = 1/4#

Explanation:

Each term in a sequence - whether geometric or arithmetic can be given as a formula or as a value.
In a GP, #a_n = ar^(n-1)#

In this example, we have #a_1 = -16 and a_4 = 1024#, which can be written as #ar^0 and ar^3#

Let's divide them - formulae and values:

#(ar^3)/(ar) = (-16)/(1024) #

#(cancelar^3)/(cancelar) = (-16)/1024 " " rArr r^3 = 1/(-64)#

Finding the cube root gives #r = 1/-4#.

Now that we have #a# and #r# we can find the value of any term in the sequence.

#a_7 = 1024 (1/-4)^(6) = 1/4#