Question

A G.P consists of an even number of terms. If the sum of all the terms is five times the sum of the terms occupying odd places, the common ratio is
a) 2
b) 3
c) 4
d) 5

Answer 1

Hint: In this question, first of all we have to calculate the sum of all the terms of a G.P. After this, we have to calculate the sum of terms occupying odd places and then we will equate the two sums with the help of the condition that is given in the question.

Complete step-by-step answer:

In general,

${\text{Let }}a,ar,a{r^2},...,a{r^{n - 1}}{\text{ be the G}}{\text{.P}}$ Where a is the first term and r is the common ratio

General formula for calculating the sum of all the terms in a G.P is denoted as Sn and is equal to

$

   \Rightarrow Sn = \dfrac{{a({r^n} - 1)}}{{r - 1}}{\text{ if (}}r > 1{\text{ and r}} < - 1) \\

   \Rightarrow Sn = \dfrac{{a(1 - {r^n})}}{{1 - r}}{\text{ if ( - 1}} < r < 1) \\

$

Let n be any number of terms

$\therefore {\text{2n will be the even number of terms of the G}}{\text{.P as mentioned in the question}}$

Now, according to the given question

Sum of all the terms = 5(Sum of the terms occupying odd places)

$ \Rightarrow A1 + A2 + A3 + ... + A2n = 5(A1 + A3 + A5 + ... + A2n - 1)$

$ \Rightarrow a + ar + a{r^2} + ... + a{r^{2n - 1}} = 5(a + a{r^2} + a{r^4} + ... + a{r^{2n - 2}})$

Now, the number of terms in the GP that is made of terms occupying odd places is n and the common ratio is ${r^2}$.

$ \Rightarrow a\{ \dfrac{{(1 - {r^{2n}})}}{{(1 - r)}}\} = 5a\{ \dfrac{{(1 - {{({r^2})}^n})}}{{(1 - {r^2})}}\} $

After simplifying, we get

\[ \Rightarrow a\{ \dfrac{{(1 - {r^{2n}})}}{{(1 - r)}}\} = 5a\{ \dfrac{{(1 - {r^2}^n)}}{{(1 - r)(1 + r)}}\} \]

$ \Rightarrow 1 + r = 5$

$ \Rightarrow r = 4$

$\therefore {\text{ Option c}}{\text{ is correct}}$

Note: In the above question, we have considered the value of r as $r > 1{\text{ }}and{\text{ }}r < - 1$. We can also take the value of r as $ - 1 < r < 1$.The result would not have been affected with that. If we don’t know the number of terms of G.P and still we want to calculate the sum of an infinite G.P, we can do that with the help of the formula $\dfrac{a}{{1 - r}}$ but the value of r should be$ - 1 < r < 1$. In case the value of r is greater than 1, then in that case, the sum would rise to infinity ($\infty $).