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.