Question

Which term of the GP. 2, 2\[\sqrt{2}\], 4,... is 128\[\sqrt{2}\]?

Answer 1

Given,

G.P. : \[2, 2\sqrt{2}, 4, ........\]

In above G.P.,

First term (a) \[{} = 2\]

Common ratio (r) \[{} = \frac{2\sqrt{2}}{2} = \sqrt{2}\]

Let \[n^{\text{th}}\] term of G.P. be \[128\sqrt{2}.\]

⇒ \[ar^{n \: - \: 1} = 128\sqrt{2}\]

⇒ \[2 \times (\sqrt{2})^{n \: - \: 1} = 128\sqrt{2}\]

⇒ \[2 \times \left(2^{\frac{1}{2}}\right)^{n \: - \: 1} = (2)^7 \cdot \sqrt{2}\]

⇒ \[2 \times (2)^{\frac{n \: - \: 1}{2}} = 2^7 \cdot 2^{\frac{1}{2}}\]

⇒ \[(2)^{1 \: + \: \frac{n \: - \: 1}{2}} = (2)^{7 \: + \: \frac{1}{2}}\]

⇒ \[(2)^{\frac{n \: - \: 1 \: + \: 2}{2}} = (2)^{\frac{14 \: + \: 1}{2}}\]

⇒ \[(2)^{\frac{n \: + \: 1}{2}} = (2)^{\frac{15}{2}}\]

⇒ \[\frac{n + 1}{2} = \frac{15}{2}\]

⇒ n + 1 = 15

⇒ n = 15 − 1 = 14

Hence, 14^th term of G.P. = 128\[\sqrt{2}\]