Question

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?

Answer 1

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`.