Solveeit Logo

Question

Question: What is the term-to-term rule to this sequence? \(64,32,16,8,4\)...

What is the term-to-term rule to this sequence? 64,32,16,8,464,32,16,8,4

Explanation

Solution

A geometric sequence is the product of an infinite number of terms with a fixed ratio between them. In this type of sequence, the previous term is multiplied by a constant to find each term. The given sequence is a geometric sequence.

Complete step by step solution:
Since we know that the above given sequence is a geometric sequence.
In general, geometric sequence is written in the given form –
a,ar,ar2,ar3,.....\\{ a,ar,a{r^2},a{r^3},.....\\}
Here, aais the first term of the sequence and rris the common ratio.
We can observe that every next term in the given sequence is obtained by multiplying with 3 to the previous term. Like this the whole sequence is formed.
Now, let’s take the sequence given in the question –
64,32,16,8,4,....64,32,16,8,4,....
Here, in the given sequence, we can see that there is a particular number by which on multiplying with the previous number, the next number is obtained.
Let’s analyse the sequence –
6432=2\dfrac{{64}}{{32}} = 2 or 642=32\dfrac{{64}}{2} = 32
3216=2\dfrac{{32}}{{16}} = 2 or 322=16\dfrac{{32}}{2} = 16
168=2\dfrac{{16}}{8} = 2 or 162=8\dfrac{{16}}{2} = 8
So we can see that in order to get 3232, we have 6464 by 22 or we can say that we have to multiply 6464 by 12\dfrac{1}{2}. Same procedure is done for all the terms in order to get its next term.
Now if we assume 6464 as the first term and call it as an{a_n}, 3232 can be written as an1{a_{n - 1}} and so on. Then an=12×an1{a_n} = \dfrac{1}{2} \times {a_{n - 1}}
So, for the term rule, an=12×an1{a_n} = \dfrac{1}{2} \times {a_{n - 1}} is the notation by which we can get the successive term of the given sequence.

Note:
Geometric sequence played a key role in the early development of calculus, and they're widely used in physics, engineering, biology, economics, computer science, queueing theory, and finance.