Question

If the sum to infinite series \[3+\dfrac{\left( 3+d \right)1}{4}+(3+2d)\dfrac{1}{{{4}^{2}}}+(3+3d)\dfrac{1}{{{4}^{3}}}+...............\] is \[\dfrac{44}{9}\] . Then, find the value of d
A.1
B.2
C.-1
D.4

Answer 1

Hint: Break the given series into summation of two series as \[S={{S}_{1}}+{{S}_{2}}\] where, \[S=3+\dfrac{\left( 3+d \right)1}{4}+(3+2d)\dfrac{1}{{{4}^{2}}}+(3+3d)\dfrac{1}{{{4}^{3}}}+...............\] , \[{{S}_{1}}=3+\dfrac{3}{4}+\dfrac{3}{{{4}^{2}}}+\dfrac{3}{{{4}^{3}}}+...............\] , and
\[{{S}_{2}}=\dfrac{d}{4}+\dfrac{2d}{{{4}^{2}}}+\dfrac{3d}{{{4}^{3}}}+...............\] . Here, we can see that \[{{S}_{1}}\] is an infinite Geometric progression series which has 3 as its first term and \[\dfrac{1}{4}\] as its common ratio. Find the summation using the formula,
\[Sum=\dfrac{First\,term}{1-common\,ratio}\] . Multiply by \[\dfrac{1}{4}\] in the series \[{{S}_{2}}\] and then subtract it from \[{{S}_{2}}\] . Then calculate \[{{S}_{2}}\] in terms of d. Now, put the values of \[{{S}_{1}}\] and \[{{S}_{2}}\] in the equation \[S={{S}_{1}}+{{S}_{2}}\] . In the question we have the value of S. Now solve it further.

Complete step-by-step answer:
According to the question, we have the series \[3+\dfrac{\left( 3+d \right)1}{4}+(3+2d)\dfrac{1}{{{4}^{2}}}+(3+3d)\dfrac{1}{{{4}^{3}}}+...............\] and the value of this series is \[\dfrac{44}{9}\] . We can see that our series is complex, so we have to make this series into simpler form.
First of all, break this series into summation of two series as,
\[S={{S}_{1}}+{{S}_{2}}\] …………………(1)
Where, \[{{S}_{1}}=3+\dfrac{3}{4}+\dfrac{3}{{{4}^{2}}}+\dfrac{3}{{{4}^{3}}}+...............\] ………………..(2),
 \[{{S}_{2}}=\dfrac{d}{4}+\dfrac{2d}{{{4}^{2}}}+\dfrac{3d}{{{4}^{3}}}+...............\] ……………………(3)
It is given that the value of the series,
\[S=3+\dfrac{\left( 3+d \right)1}{4}+(3+2d)\dfrac{1}{{{4}^{2}}}+(3+3d)\dfrac{1}{{{4}^{3}}}+...............\] = \[\dfrac{44}{9}\] ……………………………(4)
We can see that the series in equation (2) is an infinite G.P series which as 3 as its first term and
\[\dfrac{1}{4}\] as its common ratio.
We know the formula of the summation of an infinite G.P, \[Sum=\dfrac{First\,term}{1-common\,ratio}\] .
Now, using this formula to calculate the summation of the series in equation (2),
\[\begin{align}
  & {{S}_{1}}=3+\dfrac{3}{4}+\dfrac{3}{{{4}^{2}}}+\dfrac{3}{{{4}^{3}}}+............... \\
 & \Rightarrow {{S}_{1}}=\left( \dfrac{3}{1-\dfrac{1}{4}} \right) \\
 & \Rightarrow {{S}_{1}}=\left( \dfrac{3}{\dfrac{3}{4}} \right) \\
\end{align}\]
\[\Rightarrow {{S}_{1}}=4\] …………………..(5)
Now, we have to solve the series in equation (3).
Multiplying by \[\dfrac{1}{4}\] in equation (3), we get
\[{{S}_{2}}.\dfrac{1}{4}=\dfrac{d}{4}.\dfrac{1}{4}+\dfrac{2d}{{{4}^{2}}}.\dfrac{1}{4}+\dfrac{3d}{{{4}^{3}}}.\dfrac{1}{4}+...............\]
\[\Rightarrow \dfrac{{{S}_{2}}}{4}=\dfrac{d}{{{4}^{2}}}+\dfrac{2d}{{{4}^{3}}}+\dfrac{3d}{{{4}^{4}}}+...............\] ………………………….(6)
Subtracting equation (6) from equation (3), we get
\[{{S}_{2}}=\dfrac{d}{4}+\dfrac{2d}{{{4}^{2}}}+\dfrac{3d}{{{4}^{3}}}+...............\]
\[\dfrac{{{S}_{2}}}{4}=\dfrac{d}{{{4}^{2}}}+\dfrac{2d}{{{4}^{3}}}+\dfrac{3d}{{{4}^{4}}}+...............\]
\[\begin{align}
  & {{S}_{2}}-\dfrac{{{S}_{2}}}{4}=\dfrac{d}{4}+\dfrac{2d}{{{4}^{2}}}-\dfrac{d}{{{4}^{2}}}+\dfrac{3d}{{{4}^{3}}}-\dfrac{2d}{{{4}^{3}}}.................... \\
 & \Rightarrow \dfrac{3{{S}_{2}}}{4}=\dfrac{d}{4}+\dfrac{d}{{{4}^{2}}}+\dfrac{d}{{{4}^{3}}}+............. \\
\end{align}\]
We can see that the above series is an infinite G.P series which has \[\dfrac{d}{4}\] as its first term and \[\dfrac{1}{4}\] as its common ratio.
We know the formula of the summation of an infinite G.P, \[Sum=\dfrac{First\,term}{1-common\,ratio}\] .
Now, using this formula to calculate the summation of the series,
\[\begin{align}
  & \Rightarrow \dfrac{3{{S}_{2}}}{4}=\dfrac{d}{4}+\dfrac{d}{{{4}^{2}}}+\dfrac{d}{{{4}^{3}}}+............. \\
 & \Rightarrow \dfrac{3{{S}_{2}}}{4}=\dfrac{\dfrac{d}{4}}{1-\dfrac{1}{4}} \\
 & \Rightarrow \dfrac{3{{S}_{2}}}{4}=\dfrac{\dfrac{d}{4}}{\dfrac{3}{4}} \\
 & \Rightarrow \dfrac{3{{S}_{2}}}{4}=\dfrac{d}{3} \\
\end{align}\]
\[\Rightarrow {{S}_{2}}=\dfrac{4d}{9}\] ……………………….(7)
From equation (1), we have \[S={{S}_{1}}+{{S}_{2}}\] .
Now, putting the values of \[S\] , \[{{S}_{1}}\] , and \[{{S}_{2}}\] in the equation \[S={{S}_{1}}+{{S}_{2}}\] , we get
\[\begin{align}
  & S={{S}_{1}}+{{S}_{2}} \\
 & \Rightarrow \dfrac{44}{9}=4+\dfrac{4d}{9} \\
 & \Rightarrow \dfrac{44}{9}-4=\dfrac{4d}{9} \\
 & \Rightarrow \dfrac{8}{9}=\dfrac{4d}{9} \\
 & \Rightarrow 2=d \\
\end{align}\]
So, the value of common ratio is 2.
Hence, the correct option is B.

Note: In this question, one might think to add equation (3) and equation (6).
On adding, \[{{S}_{2}}=\dfrac{d}{4}+\dfrac{2d}{{{4}^{2}}}+\dfrac{3d}{{{4}^{3}}}+...............\] and \[\dfrac{{{S}_{2}}}{4}=\dfrac{d}{{{4}^{2}}}+\dfrac{2d}{{{4}^{3}}}+\dfrac{3d}{{{4}^{4}}}+...............\] we get,
\[\dfrac{5{{S}_{2}}}{4}=\dfrac{d}{4}+\dfrac{3d}{{{4}^{2}}}+\dfrac{5d}{{{4}^{3}}}+................\] .
We can see that the above series is not an infinite G.P series and we don’t have any direct formula to find the summation of these types of series. So, we need to convert it into an infinite G.P series. Therefore, we need to subtract equation (6) from equation (3).