The problem provided involves a geometric progression (or sequence) where you need to find the first term $a_1$, the common ratio $q$, and the sixth term $a_6$. Given information includes:

• $a_2 = 2$
• $a_5 = 16$

To solve this, we can use the properties of a geometric progression. The $n$-th term of a geometric sequence is given by:

$a_n = a_1 \cdot q^{n-1}$

Step 1: Express $a_2$ and $a_5$ in terms of $a_1$ and $q$

1. Since $a_2 = a_1 \cdot q$, we have: $a_1 \cdot q = 2$

2. Since $a_5 = a_1 \cdot q^4$, we have: $a_1 \cdot q^4 = 16$

Step 2: Solve for $q$ and $a_1$

We can divide the equation for $a_5$ by the equation for $a_2$ to eliminate $a_1$:

$\frac{a_1 \cdot q^4}{a_1 \cdot q} = \frac{16}{2}$

Simplifying, we get:

$q^3 = 8$

Taking the cube root of both sides, we find:

$q = 2$

Now, substitute $q = 2$ back into the equation for $a_2$:

$a_1 \cdot 2 = 2 \Rightarrow a_1 = 1$

Step 3: Find $a_6$

Now that we know $a_1 = 1$ and $q = 2$, we can find $a_6$:

$a_6 = a_1 \cdot q^5 = 1 \cdot 2^5 = 32$

Final Answer

• $a_1 = 1$
• $q = 2$
• $a_6 = 32$

Would you like a more detailed breakdown, or do you have additional questions?

Here are 5 questions to explore further:

1. What if the common ratio $q$ was negative? How would it affect the sequence?
2. Can you find a general formula for the sum of the first $n$ terms of this geometric sequence?
3. What is the sum of the first 6 terms of this sequence?
4. How would you approach this if given different terms, say $a_3$ and $a_7$, instead of $a_2$ and $a_5$?
5. What are some real-life applications of geometric progressions?

Tip: When given a geometric progression problem, try to simplify equations by dividing terms with the same base to solve for the common ratio easily.