Answer
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Hint: In a geometric progression (GP) or geometric sequence of the form $b,br,b{{r}^{2}},...,b{{r}^{n-1}}$ , we can find the common ratio by dividing the second term by the first term or dividing third term by the second term , or so on. In order to find the common ratio of the given GP, we have to divide the second term by the first term.
Complete step by step answer:
We have to find the common ratio of the geometric sequence 81, 27, 9, 3, … . We know that geometric progression (GP) or geometric sequence, is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. We can represent geometric sequence in the form $b,br,b{{r}^{2}},...,b{{r}^{n-1}}$ , where b is the first term, r is the common ratio and $b{{r}^{n-1}}$ is the ${{n}^{th}}$ term.
To find the common ratio of a geometric sequence, we have to divide the second term by the first term or we can divide the third term by the second term , or so on.
$\begin{align}
& \Rightarrow \dfrac{\text{Second term}}{\text{First term}}=\dfrac{\require{cancel}\cancel{b}r}{\require{cancel}\cancel{b}}=r \\
& \Rightarrow \dfrac{\text{Third term}}{\text{Second term}}=\dfrac{b{{r}^{2}}}{br}=r \\
\end{align}$
We are given a GP 81, 27, 9, 3, … To find the common ratio of this GP, we have to divide the second term by the first term.
$\Rightarrow r=\dfrac{\text{Second term}}{\text{First term}}$
Here, we can see that the second term is 27 and the first term is 81. Therefore, we can find the common ratio as
$\Rightarrow r=\dfrac{27}{81}$
Let us cancel the common factor 27.
$\Rightarrow r=\dfrac{1}{3}$
Therefore, the common ratio is $\dfrac{1}{3}$ .
So, the correct answer is “Option b”.
Note: Students may get confused with geometric sequence and arithmetic sequence. We usually call arithmetic sequences as arithmetic progression (AP). An AP is a mathematical sequence in which the difference between two consecutive terms is always a constant. We can represent an AP as ${{a}_{1}},{{a}_{2}},...,{{a}_{n}}$ where ${{a}_{1}}$ is the first term and ${{a}_{n}}$ is the last term. In an AP, the sequence is such that there is a common difference between the term which is found by subtracting the first term from the second term, or the second term from the third term or so on. Students may get confused with the common ratio in GP and the common difference in an AP.
Complete step by step answer:
We have to find the common ratio of the geometric sequence 81, 27, 9, 3, … . We know that geometric progression (GP) or geometric sequence, is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. We can represent geometric sequence in the form $b,br,b{{r}^{2}},...,b{{r}^{n-1}}$ , where b is the first term, r is the common ratio and $b{{r}^{n-1}}$ is the ${{n}^{th}}$ term.
To find the common ratio of a geometric sequence, we have to divide the second term by the first term or we can divide the third term by the second term , or so on.
$\begin{align}
& \Rightarrow \dfrac{\text{Second term}}{\text{First term}}=\dfrac{\require{cancel}\cancel{b}r}{\require{cancel}\cancel{b}}=r \\
& \Rightarrow \dfrac{\text{Third term}}{\text{Second term}}=\dfrac{b{{r}^{2}}}{br}=r \\
\end{align}$
We are given a GP 81, 27, 9, 3, … To find the common ratio of this GP, we have to divide the second term by the first term.
$\Rightarrow r=\dfrac{\text{Second term}}{\text{First term}}$
Here, we can see that the second term is 27 and the first term is 81. Therefore, we can find the common ratio as
$\Rightarrow r=\dfrac{27}{81}$
Let us cancel the common factor 27.
$\Rightarrow r=\dfrac{1}{3}$
Therefore, the common ratio is $\dfrac{1}{3}$ .
So, the correct answer is “Option b”.
Note: Students may get confused with geometric sequence and arithmetic sequence. We usually call arithmetic sequences as arithmetic progression (AP). An AP is a mathematical sequence in which the difference between two consecutive terms is always a constant. We can represent an AP as ${{a}_{1}},{{a}_{2}},...,{{a}_{n}}$ where ${{a}_{1}}$ is the first term and ${{a}_{n}}$ is the last term. In an AP, the sequence is such that there is a common difference between the term which is found by subtracting the first term from the second term, or the second term from the third term or so on. Students may get confused with the common ratio in GP and the common difference in an AP.
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