Question

The sum of an infinite G.P is 4 and the sum of cubes of its terms is 92. The common ratio of original G.P is
(a) \[\dfrac{1}{2}\]
(b) \[\dfrac{2}{3}\]
(c) \[\dfrac{1}{3}\]
(d) \[-\dfrac{1}{2}\]
(e) None of the above.

Answer 1

Hint: We solve this problem using the G.P formulas for infinite terms.
The general representation of G.P is given as
\[a,ar,a{{r}^{2}},a{{r}^{3}},.....\]
Where, \[a\] is the first term and \[r\] is the common ratio.
We have the formula for the sum of infinite G.P is given as
\[{{S}_{\infty }}=\dfrac{a}{1-r},\left| r \right|<1\]
By using the above formula e find the required value.

Complete step by step answer:
We are given that the sum of infinite G.P is 4
Let us assume that the given G.P as
\[a,ar,a{{r}^{2}},a{{r}^{3}},.....\]
Where, \[a\] is the first term and \[r\] is the common ratio.

We know that the formula for sum of infinite G.P is given as
\[{{S}_{\infty }}=\dfrac{a}{1-r},\left| r \right|<1\]
By using the above formula we get
\[\begin{align}
  & \Rightarrow \dfrac{a}{1-r}=4 \\
 & \Rightarrow a=4\left( 1-r \right)......equation(i) \\
\end{align}\]
We are given that the sum of cubes of its terms is 92
Now, let us cube the each term in the above assumed G.P then we get the terms of new G.P as
\[{{a}^{3}},{{a}^{3}}{{r}^{3}},{{a}^{3}}{{r}^{6}},......\]
Here, we can see that the first term is \[{{a}^{3}}\] and common ratio is \[{{r}^{3}}\]
Now, by using the infinite G.P formula to this new G.P then we get
\[\begin{align}
  & \Rightarrow \dfrac{{{a}^{3}}}{1-{{r}^{3}}}=92 \\
 & \Rightarrow {{a}^{3}}=92\left( 1-{{r}^{3}} \right) \\
\end{align}\]
Now, by substituting the value of \[a\] from equation (i) then we get
\[\begin{align}
  & \Rightarrow {{\left[ 4\left( 1-r \right) \right]}^{3}}=92\left( 1-{{r}^{3}} \right) \\
 & \Rightarrow 64{{\left( 1-r \right)}^{3}}=92\left( 1-{{r}^{3}} \right) \\
\end{align}\]
We know that the formula of difference of cubes of two numbers as
\[{{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)\]
By using this formula in above equation then we get
\[\begin{align}
  & \Rightarrow 64{{\left( 1-r \right)}^{3}}=92\left( 1-r \right)\left( 1+r+{{r}^{2}} \right) \\
 & \Rightarrow 16{{\left( 1-r \right)}^{2}}=23\left( 1+r+{{r}^{2}} \right) \\
\end{align}\]
We also know that the formula of square of difference of numbers as
\[{{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}\]

By using this formula in above equation then we get
\[\begin{align}
  & \Rightarrow 16\left( 1-2r+{{r}^{2}} \right)=23\left( 1+r+{{r}^{2}} \right) \\
 & \Rightarrow 16-32r+16{{r}^{2}}=23+23r+23{{r}^{2}} \\
 & \Rightarrow 7{{r}^{2}}+55r+7=0 \\
\end{align}\]
We know that the roots of quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By using this formula to above equation then we get
\[\begin{align}
  & \Rightarrow r=\dfrac{-55\pm \sqrt{{{55}^{2}}-4\left( 7 \right)\left( 7 \right)}}{2\left( 7 \right)} \\
 & \Rightarrow r=\dfrac{-55\pm \sqrt{3025-196}}{14} \\
 & \Rightarrow r=\dfrac{-55\pm \sqrt{2829}}{14} \\
\end{align}\]
Here, we can see that we get two values for \[r\] one for positive and other for negative.
We know that \[\left| r \right|<1\]
Here, we can see that only a positive sign gives the value of common ratio such that \[\left| r \right|<1\]
So, we can conclude that the common ratio of original G.P is
\[\therefore r=\dfrac{\sqrt{2829}-55}{14}\]
Therefore, option (e) is the correct answer.

Note:
We need to note that for the infinite G.P the common ratio should take a value between -1 and 1 that is \[\left| r \right|<1\]
If the value of the common ratio is greater than 1 or less than -1 then, the sum of the series will go on increasing forever and gives the answer as infinite.
So, the value of the common ratio for infinite G.P should be between -1 and 1.