# {4,-8,16,-32,...} In this geometric series what is the value of "q" term?

Jun 20, 2018

$4 \times - {2}^{q - 1}$

#### Explanation:

Notice that each term is equal to

$\text{the previous term} \times - 2$

By multiplying the starting number by -2 multiple times, you can find any term in the sequence.

In other words, by multiplying the starting number by some power of 2, you can find the number in the sequence that is that many terms past the starting number.

For the first term, recall that any number to the 0 power = 1. So, starting with 4, the 1st term, we can say that the first term =
$4 \times - {2}^{0} = 4 \times 1 = 4$

For the third term,
$4 \times - {2}^{2} = 4 \times 4 = 16$

Notice that the exponent we raise -2 to is one less than the number of the term we want to find: $q - 1$.

So, for the $q$th term, the equation to find it is $4 \times {2}^{q - 1}$

If you would like another example, here is how to find the 8th term:

$4 \times - {2}^{8 - 1} = 4 \times - {2}^{7} = 4 \times - 128 = - 512$