Which term of the sequence 2, 2√2, 4,... is 128? Class 9

Which term of the sequence 2, 2√2, 4,... is 128? Class 9

Question 1.

Which term of the sequence 2, $2 \sqrt{2}$, 4,... is 128? Class 9

Solution:

Here, a = 2, $r=\frac{t_2}{t_1}=\frac{2 \sqrt{2}}{2}=\sqrt{2}$

Suppose n^th term is 128,

∴ t_n = 128

$\begin{array}{ll} & 2 \times(\sqrt{2})^{n-1}=128 \\ \therefore & (\sqrt{2})^{n-1}=\frac{128}{2} \\ \therefore & (\sqrt{2})^{n-1}=64 \\ \therefore & \left(2^{\frac{1}{2}}\right)^{n-1}=2^6 \\ & 2^{\frac{n-1}{2}}=2^6 \\ \therefore & \frac{n-1}{2}=6\end{array}$

∴ n - 1 = 12

∴ n = 13

∴ 128 is the 13^th term of the sequence.