Answer

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**Hint:**

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 \[\left| r \right| < 1\].

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