Question

In a G.P of even numbers of terms, the sum of all terms is 5 times the sum of odd terms. The common ratio of the G.P is.
A) \[ - \dfrac{4}{5}\]
B) \[\dfrac{1}{5}\]
C) \[4\]
D) None of these

Answer 1

Hint:
We will use the given information and form an equation. We will solve the equation by using the geometric sum formula for a Geometric Progression to find the value of the common ratio. Geometric Progression is a set of numbers such that the consecutive numbers differ by a common ratio i.e. if we divide the second term by first we will get the same value when we divide the third term by second.

Complete Step by step Solution:
Let the number of terms in the G.P (Geometric Progression) be \[2n\].
Let its first term be \[a\] and the common ratio is \[r\].
It is given that
Sum of all the terms \[ = 5\] (sum of the terms in odd places) ……………………\[\left( 1 \right)\]
Let terms of G.P are \[{a_1},{a_2},{a_3}.........{a_{2n}}\].
Now, substituting the terms in equation \[\left( 1 \right)\] we get
\[ \Rightarrow {a_1} + {a_2} + {a_3} + {a_4}..... + {a_{2n}} = 5({a_1} + {a_3}..... + {a_{2n - 1}})\]
Substituting first term as\[a\]and all the other terms having common ratio \[r\] we can write the above equation as
  \[ \Rightarrow a + ar + a{r^2} + a{r^3} + ......a{r^{2n - 1}} = 5\left( {a + a{r^2} + a{r^4} + ..... + a{r^{2n - 2}}} \right)\]
Using geometric series sum formula \[\dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\], we get
 \[ \Rightarrow a\left( {\dfrac{{1 - {r^{2n}}}}{{1 - r}}} \right) = 5a\left( {\dfrac{{1 - {{({r^2})}^n}}}{{1 - {r^2}}}} \right)\]
Simplifying the exponent and using \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] in the denominator of RHS, we get
\[ \Rightarrow a\left( {\dfrac{{1 - {r^{2n}}}}{{1 - r}}} \right) = 5a\left( {\dfrac{{1 - {r^{2n}}}}{{(1 - r)(1 + r)}}} \right)\]
Cancelling all the common terms, we get
\[ \Rightarrow \left( {1 + r} \right) = 5\]
Subtracting 1 from both the sides, we get
\[\begin{array}{l} \Rightarrow r = 5 - 1\\ \Rightarrow r = 4\end{array}\]
So the common ratio of the G.P is 4.

Hence option (c) is correct.

Note:
As it is given that it is a G.P of even numbers we can also take it as an infinite series and use the Geometric Progression formula for infinite series.
Sum of G.P of infinite series \[ = \dfrac{a}{{1 - r}}\] where, \[a\] is the first term and\[r\] is the common ratio
So we can write the equation as
\[\left( {a + ar + a{r^2} + .......} \right) = 5\left( {a + a{r^2} + ......} \right)\]\[\begin{array}{l}\\\end{array}\]
Using the above formula, we get
\[ \Rightarrow \dfrac{a}{{1 - r}} = \dfrac{{5a}}{{1 - {r^2}}}\]
Using \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] in the denominator of RHS, we get
\[ \Rightarrow \dfrac{a}{{1 - r}} = \dfrac{{5a}}{{\left( {1 - r} \right)\left( {1 + r} \right)}}\]
On cross multiplying and cancelling out like terms, we get
\[ \Rightarrow 1 + r = 5\]
Subtracting 1 from both the sides, we get
\[\begin{array}{l} \Rightarrow r = 5 - 1\\ \Rightarrow r = 4\end{array}\]
So the common ratio of the G.P is 4.