Question:
In a G.P., the first and third terms are 4 and 8 respectively. Then the \(21^{\text{st}}\) term is
In a G.P., the first and third terms are 4 and 8 respectively. Then the \(21^{\text{st}}\) term is
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Convert powers of \( \sqrt{2} \) into powers of 2 for faster simplification.
Updated On: Apr 21, 2026
- \(4012 \)
- \(4064 \)
- \(4098 \)
- \(2048 \)
- \(4096 \)
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The Correct Option is
Solution and Explanation
Concept:
In a G.P.,
\[
a_n = a \cdot r^{n-1}
\]
Step 1: Use given terms.
\[ a = 4, \quad a_3 = ar^2 = 8 \] \[ 4r^2 = 8 \Rightarrow r^2 = 2 \Rightarrow r = \sqrt{2} \]
Step 2: Find \(21^{\text{st}}\) term.
\[ a_{21} = 4 \cdot (\sqrt{2})^{20} \] \[ (\sqrt{2})^{20} = (2)^{10} = 1024 \] \[ a_{21} = 4 \times 1024 = 4096 \]
Step 1: Use given terms.
\[ a = 4, \quad a_3 = ar^2 = 8 \] \[ 4r^2 = 8 \Rightarrow r^2 = 2 \Rightarrow r = \sqrt{2} \]
Step 2: Find \(21^{\text{st}}\) term.
\[ a_{21} = 4 \cdot (\sqrt{2})^{20} \] \[ (\sqrt{2})^{20} = (2)^{10} = 1024 \] \[ a_{21} = 4 \times 1024 = 4096 \]
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