What process do I use to find the missing numbers in a geometric sequence, when there are not two consecutive numbers? For example:81, __, __, __, 1 and 9/4, __, __, 2/3, __  

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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In geometric progression we know that:

an = a1* r^(n-1)

such that (an) is any term in the progression, a1 is the first term, and r is the common difference.

In the first example:

Given a1= 81

==> a5 = 1

Then we will need to find the common difference r , then we will calculate the other terms.

==> a5 = a1* r^4

==> 1= 81*r^4

==> r^4 = 1/81

==> r = +-1/3

Now since we have 2 values for r, then we have two possible sequences.

==> When r= 1/3

==> a1= 81

==> a2= 81*1/3  = 27

==> a3= 81*1/3^2 = 9

==> a4= 81*1/3^3 = 3

==> a5 = 81^1/3^4 = 1

==> The sequence is: 81, 27, 9, 3, 1 with a common difference r= 1/3

Now when r= -1/3

==> a1= 81

==> a2= 81*-1/3 = -27

==> a3= 81*(-1/3)^2 = 9

==> a4= 81*(-1/3)^3 = -3

==> a5= 81*(-1/3)^4 = 1

Then the sequence is: 81, -27, 9, -3, 1 with the common difference r= -1/3

 

Now for the example number 2:

a1= 9/4

a4= 2/3

We will determine r.

==> a4= a1*r^3

==> 2/3 = 9/4 * r^3

==> r^3 = 8/27

==> r= 2/3

Then the common difference is 2/3

==> a1= 9/4

==> a2= 9/4* 2/3 = 3/2

==> a3= 9/4 * (2/3)^2 = 1

==> a4= 9/4* (2/3)^3 = 2/3

==> a5= 9/4*(2/3)^4 = 4/9

Then the sequence is: 9/4, 3/2, 1, 2/3, 4/9  which common difference r= 2/3

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