Question

If in an infinite G.P first term is equal to the twice of the sum of the remaining terms, then the common ratio is
$A)1$
$B)2$
$C)\dfrac{1}{3}$
$D)\dfrac{{ - 1}}{3}$

Answer 1

Hint: First, we need to know about the concept of AM and GM. An arithmetic progression that can be given by $a,(a + d),(a + 2d),(a + 3d),...$where $a$ is the first term and $d$ is a common difference.
A geometric progression can be given by $a,ar,a{r^2},....$ where $a$ is the first term and $r$ is a common ratio.
In this question, they gave the concept of G.P, which we make use of $a,ar,a{r^2},....$ to find the required common ratio.

Complete step-by-step solution:
Since from the given that first term is equal to twice of the sum of the remaining terms, which means in the term G.P $a,ar,a{r^2},....\infty $ (infinite sequence) having the first term as $a$ which is equal to the sum of the remaining term, $ar + a{r^2} + ....\infty $
Since the sum of the n-terms can be expressed as for GP with the common ratio the formula to be calculated $GP = \dfrac{a}{{1 - r}},r \ne 1,r > 0$, where $a$ is the first term.
But here the sum of the remaining terms means $ar + a{r^2} + ....\infty $ where $ar$ is the first term.
Thus, we get twice $GP = \dfrac{{ar}}{{1 - r}}$ equals to the first term $a$
Hence, we get$a = 2(\dfrac{{ar}}{{1 - r}})$ and common values canceled $a = 2(\dfrac{{ar}}{{1 - r}}) \Rightarrow 1 = \dfrac{{2r}}{{1 - r}} \Rightarrow 1 - r = 2r$
Further solving we get $1 - r = 2r \Rightarrow r = \dfrac{1}{3}$ is the common ratio.
Therefore, the option $C)\dfrac{1}{3}$ is correct.

Note: Geometric Progression:
New series is obtained by multiplying the two consecutive terms so that they have constant factors.
In GP the series is identified with the help of a common ratio between consecutive terms.
Series vary in the exponential form because it increases by multiplying the terms.
The general GP formula for the \[{n^{th}}\] term is given as ${a_n} = a{r^{n - 1}}$