The sum of the first #5# terms of a geometric sequence is #93# and the #10#th term is #3/2#. What is the common ratio and what is the fourth term?

1 Answer
Jul 26, 2016

Answer:

If #3/2# is a typo for #3/32# then the common ratio is #1/2# and the #4#th term is #6#.

Explanation:

The general term of a geometric sequence is expressible by the formula:

#a_n = ar^(n-1)#

where #a# is the initial term and #r# the common ratio.

If #r != 1# then the sum of the first #N# terms is given by the formula:

#sum_(n=1)^N a_n = sum_(n=1)^N ar^(n-1) = a(r^N-1)/(r-1)#

In our example, we are given:

#93 = sum_(n=1)^5 a_n = a(r^5-1)/(r-1)#

#ar^9 = a_10 = 3/2#

If this is to have rational solutions, then we should look at the factors of #93#.

The prime factorisation of #93# is:

#93 = 3*31 = 3*(2^5-1)/(2-1)#

So #93# is the sum of the first five terms of the geometric sequence:

#3, 6, 12, 24, 48#

But if that were our sequence, then #a_10 = 32^9 = 1536# - somewhat larger than #3/2#.

How about if we reverse the sequence?

#48, 24, 12, 6, 3#

Then the #6#th term will be #3/2#, not the #10#th term - which will be #48*(1/2)^9 = 3/32#

Is there a typo in the question? Should the #10#th term be specified as #3/32#?

If so, then the common ratio is #1/2# and the #4#th term is #6#.

If the question is correct in the form given then the answer will be much more messy. Specifically, the common ratio #r# would be the Real zero of the nonic polynomial: #62r^9-r^4-r^3-r^2-r-1#, an irrational number approximately #0.709461#.