Question

The geometric series \[a+ar+a{{r}^{2}}+a{{r}^{3}}+...............\infty \] has sum 7 and the terms involving odd powers of r has sum ‘3’, then the value of \[\left( {{a}^{2}}-{{r}^{2}} \right)\] is
(A) \[\dfrac{5}{4}\]
(B) \[\dfrac{5}{2}\]
(C) \[\dfrac{25}{4}\]
(D) 5

Answer 1

Hint: We have two infinite geometric series, \[\left( a+ar+a{{r}^{2}}+a{{r}^{3}}+...............\infty \right)\] and \[\left( ar+a{{r}^{3}}+a{{r}^{5}}...............\infty \right)\] . Get the common ratio of both geometric series using the formula, \[\text{Common ratio = }\dfrac{\text{second term}}{\text{first term}}\] . Now, we have the summation of these two geometric series which is equal to 7 and 3 respectively. We know the formula of infinite geometric series, \[\text{Sum=}\dfrac{\text{First}\,\text{term}}{\text{1-common}\,\text{ratio}}\] . Now, we have two equations in terms of the variables a and r. Solve it further and put the value of a and r in \[\left( {{a}^{2}}-{{r}^{2}} \right)\] .

Complete step by step answer:
According to the question, it is given that the geometric series \[\left( a+ar+a{{r}^{2}}+a{{r}^{3}}+...............\infty \right)\] has sum 7 and the terms involving odd powers of r has sum ‘3’.
In case \[{{1}^{st}}\] , we have the summation of the geometric series \[\left( a+ar+a{{r}^{2}}+a{{r}^{3}}+...............\infty \right)\] equal to 7.
The first term of the geometric series = a ………………………(1)
The second term of the geometric series = ar.
The common ratio of the geometric series = \[\dfrac{\text{second term}}{\text{first term}}=\dfrac{ar}{a}=r\] …………………………(2)
The sum of the Geometric series = 7 ……………………………..(3)
We know the formula of summation of infinite Geometric series, \[\text{Sum=}\dfrac{\text{First}\,\text{term}}{\text{1-common}\,\text{ratio}}\] ………………………..(4)
From equation (1) and equation (2), we have the first term and the common ratio of the geometric series.
Now, putting the value of first term and the common ratio of the geometric series in equation (4), we get
\[\text{Sum}=\dfrac{a}{1-r}\] ……………………(5)
From equation (3), we have the sum of the geometric series.
Comparing equation (3) and equation (5), we get
 \[7=\dfrac{a}{1-r}\]
\[\Rightarrow 7\left( 1-r \right)=a\] ………………….(6)
In case \[{{2}^{nd}}\] , we have the summation of the terms having odd powers of r equal to 3.
So, our geometric series is, \[\left( ar+a{{r}^{3}}+a{{r}^{5}}...............\infty \right)\] .
The first term of the geometric series = \[ar\] ………………………(7)
The second term of the geometric series = \[a{{r}^{3}}\] .
The common ratio of the geometric series = \[\dfrac{\text{second term}}{\text{first term}}=\dfrac{a{{r}^{3}}}{ar}={{r}^{2}}\] …………………………(8)
The sum of the Geometric series = 3 ……………………………..(9)
We know the formula of summation of infinite Geometric series, \[\text{Sum=}\dfrac{\text{First}\,\text{term}}{\text{1-common}\,\text{ratio}}\] ………………………..(10)
From equation (7) and equation (8), we have the first term and the common ratio of the geometric series.
Now, putting the value of first term and the common ratio of the geometric series in equation (10), we get
\[\text{Sum}=\dfrac{ar}{1-{{r}^{2}}}\] ……………………(11)
From equation (9), we have the sum of the geometric series.
Comparing equation (9) and equation (11), we get
 \[3=\dfrac{ar}{1-{{r}^{2}}}\]
\[\Rightarrow 3\left( 1-{{r}^{2}} \right)=ar\] ………………….(12)
From equation (6), we have \[7\left( 1-r \right)=a\] .
Now, dividing equation (12) by equation (6), we get
\[\dfrac{3\left( 1-{{r}^{2}} \right)}{7\left( 1-r \right)}=\dfrac{ar}{a}\]
Now, using the formula, \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)\] to simplify the term \[\left( 1-{{r}^{2}} \right)\] in the above equation, we get
\[\begin{align}
  & \Rightarrow \dfrac{3\left( 1+r \right)\left( 1-r \right)}{7\left( 1-r \right)}=r \\
 & \Rightarrow 3\left( 1+r \right)=7r \\
 & \Rightarrow 3+3r=7r \\
 & \Rightarrow 3=7r-3r \\
 & \Rightarrow 3=4r \\
 & \Rightarrow \dfrac{3}{4}=r \\
\end{align}\]
Putting, the value of r in equation (6), we get
\[\begin{align}
  & 7\left( 1-r \right)=a \\
 & \Rightarrow 7\left( 1-\dfrac{3}{4} \right)=a \\
 & \Rightarrow \dfrac{7}{4}=a \\
\end{align}\]
So, the value of a is \[\dfrac{7}{4}\] and r is \[\dfrac{3}{4}\] .
We have to find the value of \[\left( {{a}^{2}}-{{r}^{2}} \right)\] .
Putting the value of a and r in \[\left( {{a}^{2}}-{{r}^{2}} \right)\] , we get
\[\begin{align}
  & \left( {{\left( \dfrac{7}{4} \right)}^{2}}-{{\left( \dfrac{3}{4} \right)}^{2}} \right) \\
 & =\dfrac{49}{16}-\dfrac{9}{16} \\
 & =\dfrac{49-9}{16} \\
 & =\dfrac{40}{16} \\
 & =\dfrac{5}{2} \\
\end{align}\]
Therefore, the value of \[\left( {{a}^{2}}-{{r}^{2}} \right)\] is \[\dfrac{5}{2}\] .

So, the correct answer is “Option B”.

Note: In this question, one might take the common ratio of the geometric series \[\left( ar+a{{r}^{3}}+a{{r}^{5}}...............\infty \right)\] equal to the common ratio of the geometric series \[\left( a+ar+a{{r}^{2}}+a{{r}^{3}}+...............\infty \right)\] . This is wrong. The common ratio is equal to ‘r’ for the geometric series \[\left( a+ar+a{{r}^{2}}+a{{r}^{3}}+...............\infty \right)\] . The common ratio of both Geometric series is not the same. Because both of the geometric series have different first term, second term and the common ratio. Therefore, whenever we have to get the common ratio of a geometric series, we have to divide the second term by the first term of the geometric series.