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

Jul 26, 2016

If $\frac{3}{2}$ is a typo for $\frac{3}{32}$ then the common ratio is $\frac{1}{2}$ and the $4$th term is $6$.

#### Explanation:

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

${a}_{n} = a {r}^{n - 1}$

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

If $r \ne 1$ then the sum of the first $N$ terms is given by the formula:

${\sum}_{n = 1}^{N} {a}_{n} = {\sum}_{n = 1}^{N} a {r}^{n - 1} = a \frac{{r}^{N} - 1}{r - 1}$

In our example, we are given:

$93 = {\sum}_{n = 1}^{5} {a}_{n} = a \frac{{r}^{5} - 1}{r - 1}$

$a {r}^{9} = {a}_{10} = \frac{3}{2}$

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

The prime factorisation of $93$ is:

$93 = 3 \cdot 31 = 3 \cdot \frac{{2}^{5} - 1}{2 - 1}$

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 $\frac{3}{2}$.

How about if we reverse the sequence?

$48 , 24 , 12 , 6 , 3$

Then the $6$th term will be $\frac{3}{2}$, not the $10$th term - which will be $48 \cdot {\left(\frac{1}{2}\right)}^{9} = \frac{3}{32}$

Is there a typo in the question? Should the $10$th term be specified as $\frac{3}{32}$?

If so, then the common ratio is $\frac{1}{2}$ 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: $62 {r}^{9} - {r}^{4} - {r}^{3} - {r}^{2} - r - 1$, an irrational number approximately $0.709461$.