Question

The sum of the infinite G.P is $x$and the common ratio is such that $\left| r \right| < 1$. If the first term of the G.P is 2, then which of the following is correct?
A) $- 1 < x < 1$
B) $\infty < x < 1$
C) $1 < x < \infty$
D) None of the above

Answer 1

Here we have to find the range of $x$, where $x$ is the sum of the G.P. We will first find the value of $x$ in terms of $r$ using the formula of sum of the infinite G.P. We will then find the range of $r$ to get an inequality.

Formula used:
We will use the formula of sum of infinite GP given by ${\rm{sum}} = \dfrac{a}{{1 - r}}$, where $a$ is the first term and $r$ is the common ratio.

Complete step by step solution:
It is given that-
Sum of the infinite G.P $= x$
First term of this infinite G.P is 2.
Common ratio of the infinite G.P $= r$
We will substitute the value of $a$, $r$and sum of G.P in the formula ${\rm{sum}} = \dfrac{a}{{1 - r}}$.
Therefore, we get
$x = \dfrac{2}{{1 - r}}$……..$\left( 1 \right)$
It is given that $\left| r \right| < 1$. We can also write this inequality as
$- 1 < r < 1$
Now, we will multiply $- 1$ to both sides of inequality.
We know if any negative term is multiplied to the inequality, then the sign of inequality get changed.
Therefore, the inequality becomes
$\Rightarrow 1 > - r > - 1$
Adding 1 on both sides, we get
$\Rightarrow 1 + 1 > 1 - r > 1 - 1$
Thus, the inequality will change to
$\Rightarrow 0 < 1 - r < 2$
Taking inverse of the terms, we get
$\Rightarrow \infty > \dfrac{1}{{1 - r}} > \dfrac{1}{2}$
Rewriting the inequality in standard form, we get
$\Rightarrow \dfrac{1}{2} < \dfrac{1}{{1 - r}} < \infty$
Multiplying 2 to each term, we get
$\Rightarrow 1 < \dfrac{2}{{1 - r}} < \infty$
From equation 2, we have$x = \dfrac{2}{{1 - r}}$
On substituting this value, we get the inequality:-
$\Rightarrow 1 < x < \infty$

Hence, the correct option is option C.

Note:
Here we have obtained the range of the sum of the infinite G.P. G.P means Geometric Progression and it is defined as a sequence of numbers where the ratio of any term to its previous term is a fixed number and it is called a common ratio. We need to remember that the sum of infinite G.P is $\dfrac{a}{{1 - r}}$ for $\left| r \right| < 1$, but for $\left| r \right| > 1$, the sum of infinite G.P. is $\dfrac{a}{{r - 1}}$. That is why we have used the formula for sum of infinite G.P as $\dfrac{a}{{1 - r}}$ because it is given that $\left| r \right| < 1$.