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

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.