Question

If an increasing geometric series, the sum of the \[{2^{nd}}\] and the \[{6^{th}}\] term is \[\dfrac{{25}}{2}\] . The product of the \[{3^{rd}}\] and the \[{5^{th}}\] term is \[25\]. Then what is the sum of \[{4^{th}}\], \[{6^{th}}\] and \[{8^{th}}\] terms?
A. \[35\]
B. \[30\]
C. \[26\]
D. \[32\]

Answer 1

Hint In the given problem, the sum and product of the terms in geometric series are given. By using the formula of \[{n^{th}}\] term in geometric series, we will find \[{2^{nd}}\], \[{3^{rd}}\],\[{5^{th}}\] and \[{6^{th}}\]. After that, we will put the value of terms in the given condition and make two equations. By solving these equations we will get the value of the first term and common ratio. Then again using the formula of \[{n^{th}}\] term in geometric series we will calculate \[{4^{th}}\], \[{6^{th}}\] and \[{8^{th}}\] terms.

Formula used
\[{n^{th}}\] term in geometric series: \[{a_n} = a{r^{\left( {n - 1} \right)}}\]
  Where \[a\] is the first term and \[r\] is the common ratio.
\[\left( {{a^m}} \right)\left( {{a^n}} \right) = {a^{\left( {m + n} \right)}}\]

Complete step by step solution:
Given:
The sum of the \[{2^{nd}}\] and the \[{6^{th}}\] term is \[\dfrac{{25}}{2}\].
The product of the \[{3^{rd}}\] and the \[{5^{th}}\] term is \[25\].
Let \[a,ar,a{r^2},a{r^3},....\] be the increasing geometric series. Where \[a\] is the first term and \[r\] is the common ratio.
Now use the given information and form the equations.
\[ar + a{r^5} = \dfrac{{25}}{2}\] \[.....\left( 1 \right)\]
\[\left( {a{r^2}} \right)\left( {a{r^4}} \right) = 25\] \[.....\left( 2 \right)\]
Now simplify the equation \[\left( 2 \right)\].
\[{a^2}{r^6} = 25\] [ Since \[\left( {{a^m}} \right)\left( {{a^n}} \right) = {a^{\left( {m + n} \right)}}\] ]
Take the square root of the above equation.
\[a{r^3} = 5\]
\[ \Rightarrow \]\[a = \dfrac{5}{{{r^3}}}\]
Now substitute \[a = \dfrac{5}{{{r^3}}}\] in equation \[\left( 1 \right)\].
\[\left( {\dfrac{5}{{{r^3}}}} \right)r + \left( {\dfrac{5}{{{r^3}}}} \right){r^5} = \dfrac{{25}}{2}\]
Simplify the above equation.
\[\dfrac{5}{{{r^2}}} + 5{r^2} = \dfrac{{25}}{2}\]
\[ \Rightarrow \]\[\dfrac{1}{{{r^2}}} + {r^2} = \dfrac{5}{2}\]
Substitute \[{r^2} = u\] in the above equation.
\[\dfrac{1}{u} + u = \dfrac{5}{2}\]
\[ \Rightarrow \]\[{u^2} + 1 = \dfrac{5}{2}u\]
\[ \Rightarrow \]\[2{u^2} - 5u + 2 = 0\]
Factorize the above equation.
\[2{u^2} - 4u - u + 2 = 0\]
\[ \Rightarrow \]\[2u\left( {u - 2} \right) - 1\left( {u - 2} \right) = 0\]
\[ \Rightarrow \]\[\left( {2u - 1} \right)\left( {u - 2} \right) = 0\]
\[ \Rightarrow \]\[2u - 1 = 0\] and \[u - 2 = 0\]
\[ \Rightarrow \]\[u = \dfrac{1}{2}\] and \[u = 2\]
So, \[{r^2} = \dfrac{1}{2}, 2\]
\[ \Rightarrow \]\[r = \pm \sqrt {\dfrac{1}{2}} , \pm \sqrt 2 \]
Since, the sequence is increasing.
So, \[r = \sqrt 2 \] is the only possible common ratio.
Therefore, the sum of \[{4^{th}}\], \[{6^{th}}\] and \[{8^{th}}\] terms in geometric series is
\[S = a{r^3} + a{r^5} + a{r^7}\]
\[ \Rightarrow \]\[S = a{r^3}\left( {1 + {r^2} + {r^4}} \right)\]
Substitute \[a{r^3} = 5\] and \[{r^2} = 2\] in the above equation of sum.
\[S = 5\left( {1 + 2 + {2^2}} \right)\]
\[ \Rightarrow \]\[S = 5\left( {3 + 4} \right)\]
\[ \Rightarrow \]\[S = 35\]
Hence the correct option is A.

Note: Students often do a common mistake that is they are using the formula for \[{n^{th}}\] term is \[{t_n} = a{r^n}\] which is a wrong formula. The correct formula is \[{t_n} = a{r^{n - 1}}\].