An exponential sequence (G.P.) is given by 8√2, 16√2, 32√2, … . Find the n\(^{th}\) term of the sequence
- \(8\sqrt2^n\)
-
\(2^{(n+2)}\sqrt2\)
- \(\sqrt2^{(n+3)}\)
- \(8n\sqrt2\)
The correct answer is: B
Explanation
8√2, 16√2, 32√2, ..
\(a = 8\sqrt2; r =\frac{T_2}{T_1}=\frac{16\sqrt2}{8\sqrt2}=2\)
\(T_n=ar^{n-1}\)
\(T_n=8\sqrt2 \times 2^{n-1}\)
\(T_n=2^3\times2^{n-1}\times\sqrt2\)
\(T_n=2^{3+n-1}\times\sqrt2\)
\(\therefore T_n= 2^{(n+2)}\sqrt2\)