
If \[(1-p)(1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}})=1-{{p}^{6}}\], \[p\ne 1\] then the value of \[\dfrac{p}{x}\] is
A. \[\dfrac{1}{3}\]
B. \[3\]
C. \[\dfrac{1}{2}\]
D. \[2\]
Answer
163.2k+ views
Hint: In this question, we have to find the given expression \[\dfrac{p}{x}\]. For this, we need the $p$ value which is obtained from the given expression. Here, on simplifying, we get a geometric series. So, we can calculate the sum of the six terms of the series. By comparing the obtained sum to the given expression, we get the value of $p$.
Formula Used: If the series is a geometric series, then the sum of the $n$ terms is calculated by
$\begin{align}
& {{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1};r>1 \\
& {{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r};r<1 \\
\end{align}$
where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ - Sum of the $n$ terms of the series; $n$ - Number of terms; $a$ - First term in the series; $r$ - Common ratio.
The sum of infinite G.P is calculated by
${{S}_{\infty }}=\dfrac{a}{1-r}$
Here ${{S}_{\infty }}$ is the sum of the infinite terms of the series
Complete step by step solution: Given expression is
\[(1-p)(1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}})=1-{{p}^{6}}\]
On simplifying, we get
\[\Rightarrow (1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}})=\dfrac{1-{{p}^{6}}}{1-p}\text{ }...(1)\]
Here, the series formed is a geometric series. I.e.,
\[1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}}\]
Where \[a=1;r=\dfrac{9{{x}^{2}}}{3x}=3x;n=6\]
Then, the sum of those six terms in the obtained geometric series is
\[1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}}=\dfrac{1\left( 1-{{(3x)}^{6}} \right)}{1-3x}\]
So, from (1), we get
\[\dfrac{1-{{(3x)}^{6}}}{1-3x}=\dfrac{1-{{p}^{6}}}{1-p}\]
On comparing both sides, we get
\[p=3x\]
Then, the required expression is
\[\begin{align}
& p=3x \\
& \Rightarrow \dfrac{p}{x}=3 \\
\end{align}\]
Option ‘B’ is correct
Note: Here we need to use the sum of the $n$ terms of a geometric series in the given expression. So, by comparing the L.H.S and R.H.S of the obtained expression, we get the required value.
Formula Used: If the series is a geometric series, then the sum of the $n$ terms is calculated by
$\begin{align}
& {{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1};r>1 \\
& {{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r};r<1 \\
\end{align}$
where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ - Sum of the $n$ terms of the series; $n$ - Number of terms; $a$ - First term in the series; $r$ - Common ratio.
The sum of infinite G.P is calculated by
${{S}_{\infty }}=\dfrac{a}{1-r}$
Here ${{S}_{\infty }}$ is the sum of the infinite terms of the series
Complete step by step solution: Given expression is
\[(1-p)(1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}})=1-{{p}^{6}}\]
On simplifying, we get
\[\Rightarrow (1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}})=\dfrac{1-{{p}^{6}}}{1-p}\text{ }...(1)\]
Here, the series formed is a geometric series. I.e.,
\[1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}}\]
Where \[a=1;r=\dfrac{9{{x}^{2}}}{3x}=3x;n=6\]
Then, the sum of those six terms in the obtained geometric series is
\[1+3x+9{{x}^{2}}+27{{x}^{3}}+81{{x}^{4}}+243{{x}^{5}}=\dfrac{1\left( 1-{{(3x)}^{6}} \right)}{1-3x}\]
So, from (1), we get
\[\dfrac{1-{{(3x)}^{6}}}{1-3x}=\dfrac{1-{{p}^{6}}}{1-p}\]
On comparing both sides, we get
\[p=3x\]
Then, the required expression is
\[\begin{align}
& p=3x \\
& \Rightarrow \dfrac{p}{x}=3 \\
\end{align}\]
Option ‘B’ is correct
Note: Here we need to use the sum of the $n$ terms of a geometric series in the given expression. So, by comparing the L.H.S and R.H.S of the obtained expression, we get the required value.
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