
How do you write the rule for the nth term given 2, -4, 8, -16, 32,…?
Answer
531.9k+ views
Hint: Here in this question, we have a sequence and we have to check if the sequence belongs to an arithmetic sequence or not. We check the sequence with the help of geometric sequence definition and if it is an arithmetic sequence we determine the common ratio of the sequence.
Complete step-by-step answer:
In the sequence we have three kinds of sequence namely, arithmetic sequence, geometric sequence and harmonic sequence.
In arithmetic sequence we the common difference between the two terms, In geometric sequence we the common ratio between the two terms, In harmonic sequence it is a ratio of arithmetic sequence to geometric sequence.
The general geometric progression is of the form \[a,ar,a{r^2},...\] where a is first term nth d is the common ratio. The nth term of the arithmetic progression is defined as \[{a_n} = {a_0}{r^{n - 1}}\]
Now let us consider the sequence which is given in the question 2, -4, 8, -16, 32,… here we have 5 terms. Let us find the ratio between these two consecutive numbers.
Here \[{a_1} = 4\] , \[{a_2} = 7\] , \[{a_3} = 10\] , \[{a_4} = 13\] , \[{a_5} = 16\]
Let we find
The ratio between \[{a_1}\] and \[{a_2}\] , so we have
\[\dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{ - 4}}{2} = - 2\]
The ratio between \[{a_2}\] and \[{a_3}\] , so we have
\[\dfrac{{{a_3}}}{{{a_2}}} = \dfrac{8}{{ - 4}} = - 2\]
The difference between \[{a_3}\] and \[{a_4}\] , so we have
\[\dfrac{{{a_4}}}{{{a_3}}} = \dfrac{{ - 16}}{8} = - 2\]
The difference between \[{a_4}\] and \[{a_5}\] , so we have
\[\dfrac{{{a_5}}}{{{a_4}}} = \dfrac{{32}}{{ - 16}} = - 2\]
Hence we have got the ratio same for the consecutive numbers.
Therefore the given sequence is geometric sequence
Therefore the nth term of the given sequence is \[{a_n} = 2{( - 2)^{n - 1}}\]
So, the correct answer is “ \[{a_n} = 2{( - 2)^{n - 1}}\] ”.
Note: By considering the formula of arithmetic sequence we verify the common difference which we obtained. We have to check the common difference for all the terms. Suppose if we check for the first two terms not for other terms then we may go wrong. So definition of arithmetic sequence is important to solve these kinds of problems.
Complete step-by-step answer:
In the sequence we have three kinds of sequence namely, arithmetic sequence, geometric sequence and harmonic sequence.
In arithmetic sequence we the common difference between the two terms, In geometric sequence we the common ratio between the two terms, In harmonic sequence it is a ratio of arithmetic sequence to geometric sequence.
The general geometric progression is of the form \[a,ar,a{r^2},...\] where a is first term nth d is the common ratio. The nth term of the arithmetic progression is defined as \[{a_n} = {a_0}{r^{n - 1}}\]
Now let us consider the sequence which is given in the question 2, -4, 8, -16, 32,… here we have 5 terms. Let us find the ratio between these two consecutive numbers.
Here \[{a_1} = 4\] , \[{a_2} = 7\] , \[{a_3} = 10\] , \[{a_4} = 13\] , \[{a_5} = 16\]
Let we find
The ratio between \[{a_1}\] and \[{a_2}\] , so we have
\[\dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{ - 4}}{2} = - 2\]
The ratio between \[{a_2}\] and \[{a_3}\] , so we have
\[\dfrac{{{a_3}}}{{{a_2}}} = \dfrac{8}{{ - 4}} = - 2\]
The difference between \[{a_3}\] and \[{a_4}\] , so we have
\[\dfrac{{{a_4}}}{{{a_3}}} = \dfrac{{ - 16}}{8} = - 2\]
The difference between \[{a_4}\] and \[{a_5}\] , so we have
\[\dfrac{{{a_5}}}{{{a_4}}} = \dfrac{{32}}{{ - 16}} = - 2\]
Hence we have got the ratio same for the consecutive numbers.
Therefore the given sequence is geometric sequence
Therefore the nth term of the given sequence is \[{a_n} = 2{( - 2)^{n - 1}}\]
So, the correct answer is “ \[{a_n} = 2{( - 2)^{n - 1}}\] ”.
Note: By considering the formula of arithmetic sequence we verify the common difference which we obtained. We have to check the common difference for all the terms. Suppose if we check for the first two terms not for other terms then we may go wrong. So definition of arithmetic sequence is important to solve these kinds of problems.
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