How to Determine Values of x in a Geometric Progression (GP)?

Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio. Understanding how to determine the values of $x$ in a GP can be crucial for solving various mathematical and real-world problems. Let’s dive into the details.

Basic Definition and Formula

A GP can be represented as:

$a, ar, ar^2, ar^3, text{…}$

Here, $a$ is the first term, and $r$ is the common ratio. To find the $n$-th term of a GP, we use the formula:

$T_n = ar^{n-1}$

Example 1: Finding the Common Ratio

Suppose we have a GP: $2, 6, 18, 54, text{…}$

To find the common ratio ($r$), we divide the second term by the first term:

$r = frac{6}{2} = 3$

We can verify this by dividing the third term by the second term:

$r = frac{18}{6} = 3$

Example 2: Finding the $n$-th Term

If we want to find the 5th term of the GP $2, 6, 18, 54, text{…}$, we use the formula for the $n$-th term:

$T_5 = 2 times 3^{5-1} = 2 times 3^4 = 2 times 81 = 162$

Finding the Values of x Given Certain Conditions

Case 1: When Terms are Given

Consider a GP where the first term is $a$ and the common ratio is $r$. Suppose we are given the first three terms as $2, x, 18$. We need to find the value of $x$

Since $x$ is the second term, it can be represented as:

$x = ar$

We also know the third term is $18$, which can be represented as:

$18 = ar^2$

Given $a = 2$, we substitute $a$ into the equations:

$x = 2r$

$18 = 2r^2$

Solving the second equation for $r$:

$r^2 = frac{18}{2} = 9$

$r = 3 text{ or } r = -3$

Substituting $r$ back into the first equation to find $x$:

For $r = 3$:

$x = 2 times 3 = 6$

For $r = -3$:

$x = 2 times (-3) = -6$

Therefore, $x$ can be $6$ or $-6$

Case 2: When the Sum of Terms is Given

Suppose we know the sum of the first $n$ terms of a GP. The formula for the sum of the first $n$ terms ($S_n$) is:

$S_n = a frac{r^n – 1}{r – 1} text{ for } r
eq 1$

Let’s find $x$ if the sum of the first three terms of a GP is $14$, and the first term is $2$

The sum of the first three terms is:

$S_3 = a + ar + ar^2$

Given $a = 2$ and $S_3 = 14$:

$2 + 2r + 2r^2 = 14$

Dividing the entire equation by $2$:

$1 + r + r^2 = 7$

Rearranging the equation:

$r^2 + r – 6 = 0$

Factoring the quadratic equation:

$(r – 2)(r + 3) = 0$

Therefore, $r = 2$ or $r = -3$

Substituting back to find $x$:

For $r = 2$:

$x = 2r = 2 times 2 = 4$

For $r = -3$:

$x = 2r = 2 times (-3) = -6$

Thus, $x$ can be $4$ or $-6$

Solving Real-World Problems Using GP

Example: Population Growth

Consider a bacteria population that triples every hour. If the initial population is $100$, what will be the population after 5 hours?

This is a GP where $a = 100$ and $r = 3$. We want to find the 6th term $(T_6)$ because we start counting from the initial population.

Using the $n$-th term formula:

$T_6 = 100 times 3^{6-1} = 100 times 3^5 = 100 times 243 = 24300$

So, the population after 5 hours will be $24,300$

Conclusion

Determining the values of $x$ in a GP involves understanding the basic properties and formulas of geometric progressions. By using these principles, you can solve various mathematical and real-world problems efficiently. Remember to always check your work for consistency and accuracy.

Citations

  1. 1. Khan Academy – Geometric Sequences
  2. 2. Math is Fun – Geometric Progression
  3. 3. Paul’s Online Math Notes – Geometric Sequences

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ