Respuesta :
Geometric series have each term constant multiple of previous term, that constant being same for a series. From the given options, geometric series is: Option 6: -256 +64 - 16 + 4 - 1
What is a geometric sequence and how to find its nth terms?
There are three parameters which differentiate between which geometric sequence we're talking about.
The first parameter is the initial value of the sequence.
The second parameter is the quantity by which we multiply previous term to get the next term.
The third parameter is the length of the sequence. It can be finite or infinite.
Suppose the initial term of a geometric sequence is [tex]a[/tex] and the term by which we multiply the previous term to get the next term is [tex]d[/tex]
Then the sequence would look like
[tex]a, ad, ad^2, ad^3, ad^4,...[/tex]
(till the terms to which it is defined)
Thus, the nth term of such sequence would be
[tex]T_{n} = ad^{n-1}[/tex]
And therefore, we get:
[tex]\dfrac{T_{n+1}}{T_{n}} = d[/tex] (where fraction was of n+1th term to nth term of geometric series). (for all n, n being positive integer))
What is a geometric series?
When all the terms of a geometric sequence are added, then that expression is called geometric series.
Checking all options to see which one is geometric series:
- Case 1: 1
Yes it is since here we can assume n = 1, and a = 1, and d can be assumed anything.
- Case 2 and Case 3 are not series but sequence, no matter which sequence they are, they are not series, so can't be geometric series either.
- Case 4: 2 +5 + 8 + 11 + 14 + 17
If we take a = 2, then, we get 'd' as:
[tex]T_2/T_1 = d = 5/2 = 2.5[/tex]
If the given series is geometric series, then this constant d must be same for all adjacent pair of terms.
Checking for second pair, we get:
[tex]\dfrac{T_3}{T_2}= \dfrac{8}{5} = 1.6 \neq 2.5[/tex] Thus, constant wasn't preserved as the ratio is different, thus, it is not a geometric series.
- Case 5: 1+1+2+3+5+ 8 +13 +21
We see first two terms are same, which shows d must be 1, but then we see other terms are different, telling that 'd' must not be 1. Thus, d is not constant for whole series, so its not a geometric series.
- Case 6: -256 +64 - 16 + 4 - 1
If we take a = -256, then, we get 'd' as:
[tex]T_2/T_1 = d = -256/64 = -4[/tex]
If this series is geometric series(with n = 5), then the series should look like:
[tex]a+ad+ad^2+ad^3+ad^4 = -256 + \dfrac{-256}{-4} + \dfrac{-256}{16} + \dfrac{-256}{-64} + \dfrac{-256}{265}\\a+ad+ad^2+ad^3+ad^4 = -256 + 64 - 16 + 4 - 1[/tex]
The given series looks exactly same, so following the properties of a geometric series. Thus, this series is a geometric series.
Thus, From the given options, geometric series is: Option 6: -256 +64 - 16 + 4 - 1
Learn more about geometric series here:
https://brainly.com/question/1607203
