How towrite out the first five terms of each geometric sequence here? (1) a_1 = 2, r = 4 (2) a_1 = 48, r = -1/3

1 Answer
Jul 3, 2018

1) The first 5 terms are 2, 8, 32, 128, 512.
2) The first 5 terms are 48, -16, 16/3, -16/9, 16/27.

Explanation:

1) The general form of for the nth of a geometric sequence with first term a_1, common ratio between all terms r is given by

a_n=a_1(r)^(n-1)

Given a_1=2, r=4, we plug these values into the above formula, yielding a general form of the geometric sequence which we can use to get any term(s) we want:

a_n=2(4)^(n-1)

We know the first term a_1=2. That has been given.

For the second term, plug in n=2 into a_n=2(4)^(n-1):

a_2=2(4)^(2-1)
=2(4)^1=2(4)
=8

Repeat this process for the third, fourth, and fifth terms:

a_3=2(4)^(3-1)
=2(4)^2
=2(16)
=32

a_4=2(4)^(4-1)
=2(4)^3
=2(64)
=128

a_5=2(4)^(5-1)
=2(4)^4
=2(256)
=512

Thus, the first 5 terms are 2, 8, 32, 128, 512.

2) Again, plug in a_1=48, r=-1/3 into a_n=a_1(r)^(n-1), which gives us

a_n=48(-1/3)^(n-1)

We know the first term a_1=48. To determine the second, third, fourth, and fifth terms, simply plug in n=2, n=3, n=4, n=5 into the above formula:

a_2=48(-1/3)^(2-1)
=48(-1/3)=-16

a_3=48(-1/3)^(3-1)
=48(-1/3)^2
=48/9=16/3

a_4=48(-1/3)^(4-1)
=48(-1/3)^3
=-48/27=-16/9

a_5=48(-1/3)^(5-1)
=48(-1/3)^4
=48/81=16/27

Thus, the first 5 terms are 48, -16, 16/3, -16/9, 16/27.