Identify whether the following sequence is a geometric sequence or not:
.$ - 3,9, - 27,81$
Answer
611.1k+ views
Hint: We know that in the geometric sequence the ratio of the two corresponding numbers must be equal. We need to know that the ratio is equal. So, by using this we will find the ratios for the terms in the geometric progression.
Complete step by step solution:
We know that the geometric sequence is also called the geometric progression or in the short form we write it as GP to represent the geometric sequence. It is the sequence of the numbers in which the ratio of the two adjacent terms will always be constant. For example: Let us take ${a_0},{a_1},{a_2},{a_3},........$ are in GP then we can say that the ratio of the adjacent terms will be constant which means
$\dfrac{{{a_1}}}{{{a_0}}},\dfrac{{{a_2}}}{{{a_1}}},\dfrac{{{a_3}}}{{{a_2}}},$ are all equal that means that $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$
So if $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$ is given then we can say that ${a_0},{a_1},{a_2},{a_3},........$ all are in GP.
In other ways we can say that the sequence in which the second term is formed by the multiplication of the previous term by the constant term then the sequence is said to be in GP.
So let ${a_0},{a_1},{a_2},{a_3} = $$ - 3,9, - 27,81$
Now we know that $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$
So we know that
${a_0} = - 3,{a_1} = 9,{a_2} = - 27\,and\,{a_3} = 81$
So by putting the values in $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$
We get that
$ \Rightarrow \dfrac{{{a_1}}}{{{a_0}}} = \dfrac{9}{{ - 3}} = - 3,\,\,\dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{ - 27}}{9} = - 3,\,\,\dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{81}}{{ - 27}} = - 3$
Hence, we can see that $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$
So, we can say that these terms are in the geometric sequence.
Note: If the two numbers are given to be a and b then for its geometric mean the formula is given as $GM = \sqrt {ab} $
So, we must know about the formula of the sequences and their sums of the terms in order to solve such problems.
Complete step by step solution:
We know that the geometric sequence is also called the geometric progression or in the short form we write it as GP to represent the geometric sequence. It is the sequence of the numbers in which the ratio of the two adjacent terms will always be constant. For example: Let us take ${a_0},{a_1},{a_2},{a_3},........$ are in GP then we can say that the ratio of the adjacent terms will be constant which means
$\dfrac{{{a_1}}}{{{a_0}}},\dfrac{{{a_2}}}{{{a_1}}},\dfrac{{{a_3}}}{{{a_2}}},$ are all equal that means that $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$
So if $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$ is given then we can say that ${a_0},{a_1},{a_2},{a_3},........$ all are in GP.
In other ways we can say that the sequence in which the second term is formed by the multiplication of the previous term by the constant term then the sequence is said to be in GP.
So let ${a_0},{a_1},{a_2},{a_3} = $$ - 3,9, - 27,81$
Now we know that $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$
So we know that
${a_0} = - 3,{a_1} = 9,{a_2} = - 27\,and\,{a_3} = 81$
So by putting the values in $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$
We get that
$ \Rightarrow \dfrac{{{a_1}}}{{{a_0}}} = \dfrac{9}{{ - 3}} = - 3,\,\,\dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{ - 27}}{9} = - 3,\,\,\dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{81}}{{ - 27}} = - 3$
Hence, we can see that $\dfrac{{{a_1}}}{{{a_0}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}}$
So, we can say that these terms are in the geometric sequence.
Note: If the two numbers are given to be a and b then for its geometric mean the formula is given as $GM = \sqrt {ab} $
So, we must know about the formula of the sequences and their sums of the terms in order to solve such problems.
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