Question

In a G.P., the first and third terms are 4 and 8 respectively. Then the \(21^{\text{st}}\) term is

• A. \(4012 \)
• B. \(4064 \)
• C. \(4098 \)
• D. \(2048 \)
• E. \(4096 \)

Hint:

Convert powers of \( \sqrt{2} \) into powers of 2 for faster simplification.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The problem involves a Geometric Progression (G.P.), where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio `r`.
Step 2: Key Formula or Approach:
The formula for the n\(^{th}\) term of a G.P. is `a_n = a \cdot r^{(n-1)}`, where `a` is the first term and `r` is the common ratio.
1. Use the given terms to find the common ratio `r`.
2. Use the formula to find the 21\(^{st}\) term.
Step 3: Detailed Explanation:
We are given:
The first term, `a = a_1 = 4`.
The third term, `a_3 = 8`.
Using the formula for the third term:
\[ a_3 = a \cdot r^{(3-1)} = a \cdot r^2 \] Substitute the given values:
\[ 8 = 4 \cdot r^2 \] Solve for `r^2`:
\[ r^2 = \frac{8}{4} = 2 \] Now, we need to find the 21\(^{st}\) term, `a_{21}`:
\[ a_{21} = a \cdot r^{(21-1)} = a \cdot r^{20} \] We can write `r^{20}` as `(r^2)^{10}` to use our calculated value of `r^2`.
\[ a_{21} = 4 \cdot (r^2)^{10} = 4 \cdot (2)^{10} \] We know that `2^{10} = 1024`.
\[ a_{21} = 4 \cdot 1024 = 4096 \] Step 4: Final Answer:
The 21\(^{st}\) term is 4096.