We are given the following information about a geometric progression:

• The first term is $a$,
• The common ratio is $r$,
• The sum of the first two terms is 12:
  $a + ar = 12$
• The third term is 16:
  $ar^2 = 16$
• Also, we know $a > ar$ (since the first term is larger than the second term).

Let's solve this step by step:

Step 1: Express $a$ in terms of $r$

From the sum of the first two terms: $a(1 + r) = 12$ This gives: $a = \frac{12}{1 + r}$

Step 2: Substitute $a$ into the equation for the third term

Using the third term equation $ar^2 = 16$ and substituting $a = \frac{12}{1 + r}$: $\frac{12}{1 + r} \cdot r^2 = 16$ Simplifying this equation: $\frac{12r^2}{1 + r} = 16$ Multiply both sides by $1 + r$ to eliminate the denominator: $12r^2 = 16(1 + r)$ Expanding the right-hand side: $12r^2 = 16 + 16r$ Rearranging this into a standard quadratic form: $12r^2 - 16r - 16 = 0$ Simplifying by dividing through by 4: $3r^2 - 4r - 4 = 0$

Step 3: Solve the quadratic equation

We can solve $3r^2 - 4r - 4 = 0$ using the quadratic formula: $r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-4)}}{2(3)}$ $r = \frac{4 \pm \sqrt{16 + 48}}{6}$ $r = \frac{4 \pm \sqrt{64}}{6}$ $r = \frac{4 \pm 8}{6}$ So, the two possible solutions for $r$ are: $r = \frac{4 + 8}{6} = 2 \quad \text{or} \quad r = \frac{4 - 8}{6} = -\frac{2}{3}$

Step 4: Determine the correct value of $r$

Since $a > ar$, this implies $1 > r$. Therefore, the correct value of $r$ is $-\frac{2}{3}$.

Final Answer:

The value of $r$ is $-\frac{2}{3}$.

Do you want more details on any step or have any questions?

Here are five related questions:

1. How do you find the sum of a finite geometric series?
2. What happens if the common ratio is greater than 1?
3. How does the sum of an infinite geometric progression differ?
4. What are the conditions for convergence in an infinite geometric progression?
5. How do you derive the nth term of a geometric sequence?

Tip: Always double-check the signs when working with quadratic equations!