Question

The first term of an infinite geometric progression is $x$ and its sum is 5, then
A.$x < \- 10$
B.$0 < x < 10$
C.$- 10 < x < 10$
D.$x > 10$

Answer 1

Hint: The formula to calculate the sum of an infinite geometric progression is ${S_\infty } = \dfrac{a}{{1 - r}}$, where $a$ is the first term and $r$ is the common ratio of an geometric progression. Apply this formula, and then use the given conditions to find the required value. Then use $\left| r \right| < 1$ to find the inequality.

Complete step-by-step answer:
Given that the sum of an infinite geometric progression is 5 and the first term of the infinite geometric progression is $x$.
We know that the formula to calculate the sum of an infinite geometric progression is ${S_\infty } = \dfrac{a}{{1 - r}}$, where $a$ is the first term and $r$ is the common ratio of an geometric progression.
Since we know that the first term is $x$, so $a = x$.
Also, we know that the sum of an infinite geometric progression, ${S_\infty } = 5$.
Substituting these values of ${S_\infty }$ and $r$ in the above formula of sum of an infinite progression, we get
$5 = \dfrac{x}{{1 - r}}$
Multiplying the above equation by $1 - r$ on each of the sides, we get

$$\Rightarrow 5\left( {1 - r} \right) = \left( {1 - r} \right)\left( {\dfrac{x}{{1 - r}}} \right) \\\ \Rightarrow 5\left( {1 - r} \right) = x \\\$$

Dividing the above equation by 5 on each of the sides, we get

$$\Rightarrow \dfrac{{5\left( {1 - r} \right)}}{5} = \dfrac{x}{5} \\\ \Rightarrow 1 - r = \dfrac{x}{5} \\\$$

Adding the above equation by $r$ on each of the sides, we get

$$\Rightarrow 1 - r + r = \dfrac{x}{5} + r \\\ \Rightarrow 1 = \dfrac{x}{5} + r \\\$$

Subtracting the above equation by $\dfrac{x}{5}$ on each side, we get

$$\Rightarrow 1 - \dfrac{x}{5} = \dfrac{x}{5} + r - \dfrac{x}{5} \\\ \Rightarrow 1 - \dfrac{x}{5} = r \\\ \Rightarrow r = 1 - \dfrac{x}{5} \\\$$

Substituting this value of $r$ in the condition of infinite geometric progression $\left| r \right| < 1$, we get

$$\left| {1 - \dfrac{x}{5}} \right| < 1 \\\ \Rightarrow - 1 < 1 - \dfrac{x}{5} < 1 \\\$$

Subtracting the above inequality by 1 on each of the sides, we get

$$\Rightarrow - 1 - 1 < 1 - \dfrac{x}{5} - 1 < 1 - 1 \\\ \Rightarrow - 2 < \- \dfrac{x}{5} < 0 \\\$$

Multiplying the above inequality by $- 5$ on each of the sides, we get

$$\Rightarrow 10 > x > 0 \\\ \Rightarrow 0 < x < 10 \\\$$

Hence, the option B will be correct.

Note: In solving these types of questions, we should know that a series is geometric progression if and only if the common ratio of that series remains constant throughout the series. Students should remember the formulae for calculating the sum of an infinite progression, ${S_\infty } = \dfrac{a}{{1 - r}}$, where $a$ is the first term and $r$ is the common ratio of an geometric progression.