How do you find the $ 10th $ term in the geometric sequence $ 320,160,80,40,... $ ?

Hint: We know that Geometric Progression is a type of sequence where each succeeding term is produced by multiplying each consecutive term by a fixed number, which is known as a common ratio. This progression is also known as a geometric sequence of numbers that follow a pattern. Therefore, we will first find the common ratio and then apply the formula for finding the $ nth $ term of the geometric sequence.

Formulas used: $ r = \dfrac{{{a_2}}}{{{a_1}}} $ , where, $ r $ is the common ratio, $ {a_2} $ is the second term of the geometric sequence and $ {a_1} $ is the first term of the geometric sequence
 $ {a_n} = {a_1}{r^{n - 1}} $ , where, $ {a_n} $ is the $ nth $ term of the geometric sequence, $ {a_1} $ is the first term of the geometric sequence and $ r $ is the common ratio

Complete step-by-step solution:
Here, we are given the geometric series $ 320,160,80,40,... $
We have the first term $ {a_1} = 320 $ and the second term $ {a_2} = 160 $ .
Now, we will find the common ratio.
 $ r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{160}}{{320}} = \dfrac{1}{2} $
We know that the formula for finding the $ nth $ term of the geometric sequence is $ {a_n} = {a_1}{r^{n - 1}} $ .
Therefore, for finding the $ 10th $ term, we will use the formula
 $ {a_{10}} = 320 \times {\left( {\dfrac{1}{2}} \right)^{10 - 1}} = 320 \times {\left( {\dfrac{1}{2}} \right)^9} = 320 \times \left( {\dfrac{1}{{{2^9}}}} \right) $
We can write $ 320 = 10 \times 32 = 10 \times {2^5} $
\[{a_{10}} = 10 \times {2^5} \times \left( {\dfrac{1}{{{2^9}}}} \right) = 10 \times \dfrac{1}{{{2^{9 - 5}}}} = 10 \times \dfrac{1}{{{2^4}}} = \dfrac{{10}}{{16}} = \dfrac{5}{8}\]
Thus, the $ 10th $ term in the geometric sequence $ 320,160,80,40,... $ is $ \dfrac{5}{8} $ .

Note: We have seen that a geometric progression or a geometric sequence in which each term is varied by another by a common ratio. The next term of the sequence is produced when we multiply a constant which is a non-zero value, to the preceding term. In other words, we can also say that when we divide any succeeding term from its preceding term, then we get the value equal to common ratio.