Question 216 of 480
The nth term of two sequences are Qn = 3 . 2n - 2 and Um = 3 . 22m - 3. Find the product of Q2 and U2.
Correct Answer:
A
Explanation
To solve the problem, we need to find the values of \( Q_2 \) and \( U_2 \) from the given sequences and then calculate their product. Let's break this down step-by-step.
### Step 1: Calculate \( Q_2 \)
The formula for the sequence \( Q_n \) is given as:
\[
Q_n = 3 \cdot 2^{n - 2}
\]
To find \( Q_2 \), we substitute \( n = 2 \):
\[
Q_2 = 3 \cdot 2^{2 - 2}
\]
Calculating the exponent:
\[
Q_2 = 3 \cdot 2^{0}
\]
Since \( 2^0 = 1 \):
\[
Q_2 = 3 \cdot 1 = 3
\]
### Step 2: Calculate \( U_2 \)
The formula for the sequence \( U_m \) is given as:
\[
U_m = 3 \cdot 2^{2m - 3}
\]
To find \( U_2 \), we substitute \( m = 2 \):
\[
U_2 = 3 \cdot 2^{2 \cdot 2 - 3}
\]
Calculating the exponent:
\[
U_2 = 3 \cdot 2^{4 - 3} = 3 \cdot 2^{1}
\]
Since \( 2^1 = 2 \):
\[
U_2 = 3 \cdot 2 = 6
\]
### Step 3: Calculate the Product \( Q_2 \cdot U_2 \)
Now that we have both \( Q_2 \) and \( U_2 \):
\[
Q_2 = 3 \quad \text{and} \quad U_2 = 6
\]
We can find the product:
\[
Q_2 \cdot U_2 = 3 \cdot 6 = 18
\]
### Conclusion
The product of \( Q_2 \) and \( U_2 \) is \( 18 \).
### Explanation of Options
- **Option A: 18** - This is the correct answer as we calculated \( Q_2 \cdot U_2 = 18 \).
- **Option B: 12** - This is incorrect. It may arise from a miscalculation, perhaps by incorrectly calculating \( U_2 \) or \( Q_2 \).
- **Option C: 6** - This is incorrect. It could be a misunderstanding of the product of the two terms, possibly thinking only one of the terms was needed.
- **Option D: 3** - This is incorrect. It seems to be the value of \( Q_2 \) alone, not the product.
### Revision Summary
- The nth term of the sequences are calculated by substituting the respective values into the formulas.
- For \( Q_2 \), we found \( 3 \) and for \( U_2 \), we found \( 6 \).
- The product \( Q_2 \cdot U_2 \) is \( 18 \).
- Always double-check calculations to avoid common pitfalls in exponentiation and multiplication.