Question

If the sum of infinite G.P. \[p,1,\dfrac{1}{p},\dfrac{1}{{{p}^{2}}},.....,\ is\ \dfrac{9}{2}\]. Then find the value of p.

Answer 1

Hint: We will start by using the fact that the sum of an infinite G.P. with first term a and common ratio r is $\dfrac{a}{1-r}$. Then using this we will find the sum of given G.P. and equate it to $\dfrac{9}{2}$. So, that the value of p can be obtained from the equation formed.

Complete step-by-step answer:
Now, we have been given that the sum of infinite G.P.,
\[p,1,\dfrac{1}{p},\dfrac{1}{{{p}^{2}}},.....,\ is\ \dfrac{9}{2}\]
Now, we know that if an infinite G.P is,
$a,ar,a{{r}^{2}}.........$
Where a is the first term and r is a common ratio. So, we have the sum of G.P as $\dfrac{a}{1-r}$.
Now, we have the first term of G.P as p and the common ratio is $\dfrac{1}{p}$. So, we have the sum as $\dfrac{p}{1-\dfrac{1}{p}}$.
Now, we have been given it to be equal to $\dfrac{9}{2}$.
$\begin{align}
  & \dfrac{p}{1-\dfrac{1}{p}}=\dfrac{9}{2} \\
 & \dfrac{p}{\dfrac{p-1}{p}}=\dfrac{9}{2} \\
 & \dfrac{{{p}^{2}}}{p-1}=\dfrac{9}{2} \\
\end{align}$
Now, on cross – multiply we have,
$\begin{align}
  & 2{{p}^{2}}=9p-9 \\
 & 2{{p}^{2}}-9p+9=0 \\
\end{align}$
Now, using factorization method we have,
$\begin{align}
  & 2{{p}^{2}}-6p-3p+9=0 \\
 & 2p\left( p-3 \right)-3\left( p-3 \right)=0 \\
 & \left( 2p-3 \right)\left( p-3 \right)=0 \\
 & either\ 2p-3=0\ or\ p-3=0 \\
 & p=\dfrac{3}{2},p=3 \\
\end{align}$
Hence, the value of p is $\dfrac{3}{2},3$.

Note: It is important to note that we have used factorization methods to solve the quadratic equation. In this method we split the middle term as a sum of factors of the product of coefficients of ${{x}^{2}}$ and constant term.