How do you write a rule for the nth term of the geometric term given the two terms #a_3=24, a_5=96#?

1 Answer
Aug 5, 2017

Answer:

#a_n = 6 * 2^(n-1)# or #a_n = 6 * (-2)^(n-1)#

Explanation:

The general formula for a geometric sequence is #a_n = a_1 * r^(n-1)#, where #a_n# is the #n^(th)# term, #a_1# is the first term, and #r# is the common ratio.

I'm going to explain how to do this problem two ways.

The Long Way

Since we are given #a_3 = 24# and #a_5 = 96#, we can substitute them into the formula.

#a_3 = a_1 * r^(3-1) #
# a_3 = a_1 * r^2#
#color(blue)(24 = a_1 * r^2)#

#a_5 = a_1 * r^(5-1) #
# a_5 = a_1 * r^4#
#color(blue)(96= a_1 * r^4)#

Now we can solve the system of equations:

#color(blue)(24 = a_1 * r^2)# #-># solve for #a_1#

#a_1=24/r^2#

#color(blue)(96= a_1 * r^4)#

#96=24/r^2 * r^4# #-># substitute the value of #a_1# into the second equation

#96=24 * r^2#

#4=r^2#

#r=+-2#

Now that we have the value of #r#, we can find the value of #a_1#. Using the first equation, #color(blue)(24 = a_1 * r^2)#, we get

#24 = a_1 * r^2#

#24 = a_1 * (+-2)^2#

#24 = a_1 * 4#

#a_1=6#

So our formula for the sequence can be either #color(red)(a_n = 6 * 2^(n-1))# or #color(red)(a_n = 6 * (-2)^(n-1))#.

To verify if these are correct, you can write out the first few terms and see if they match the information given in the problem.

#color(red)(a_n = 6 * 2^(n-1)) #

The common ratio is #2#, so start with #6# and multiply each term by #2 => 6, 12, 24, 48, 96#

#color(red)(a_n = 6 * (-2)^(n-1))#

The common ratio is #-2#, so start with #6# and multiply each term by #-2 => 6, -12, 24, -48, 96#

In both of these formulas, #a_3=24# and #a_5=96#.

The Short Way

We are given #a_3# and #a_5#, so we can easily find out #a_4# in order to get the value of #r#.

#a_3, a_4, a_5#

#24, a_4, 96#

To find #a_4#, we can simply calculate the geometric mean.

#(a_4)^2 = 24 * 96 => a_4 = +-sqrt(24 * 96) = +-sqrt2304 = +-48#

So the three terms are either #24, 48, 96#, meaning that #r = 48/24 = 2#, or the terms are #24, -48, 96#, meaning that #r=-48/24 = -2#.

After you find #r#, you can find #a_1# the same way we did above. In the end, you get #color(red)(a_n = 6 * 2^(n-1))# or #color(red)(a_n = 6 * (-2)^(n-1))#.

(This method is easier in the context of this problem, but if you were given terms such as #a_10# and #a_19#, you would definitely want to use the first method.)