Correct Answer: Option B

Explanation

8√2, 16√2, 32√2, ..

\(a = 8\sqrt2; r =\frac{T_2}{T_1}=\frac{16\sqrt2}{8\sqrt2}=2\)

\(T_n=ar^{n-1}\)

\(T_n=8\sqrt2 \times 2^{n-1}\)

\(T_n=2^3\times2^{n-1}\times\sqrt2\)

\(T_n=2^{3+n-1}\times\sqrt2\)

\(\therefore T_n= 2^{(n+2)}\sqrt2\)

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