Show that the progression 2, $2\sqrt{2}, 4, 4\sqrt{2}$,….. is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 11th term.

Given,

2, $2\sqrt{2}, 4, 4\sqrt{2}$,…..

$\Rightarrow \dfrac{2\sqrt{2}}{2} = \dfrac{4\sqrt{2}}{4} = \sqrt{2}.$

Since, ratio between consecutive terms are equal, thus the series is in G.P.

a = 2

r = $\dfrac{2\sqrt{2}}{2} = \sqrt{2}$

We know that,

nth term of a G.P. is given by,

$\Rightarrow T$n = ar^{n - 1} \\[1em] \Rightarrow Tn = 2.( \sqrt {2})^{n - 1} \\[1em] = (\sqrt{2})^2.(\sqrt {2})^{n - 1} \\[1em] = (\sqrt{2})^{2 + n - 1} \\[1em] = (\sqrt2)^{n + 1}. \\[1em] \Rightarrow T_{11} = (\sqrt2)^{11 + 1} \\[1em] = (\sqrt2)^{12} \\[1em] = (2)^{\dfrac{1}{2} \times 12} \\[1em] = 2^{6} \\[1em] = 64.

Hence, a = 2, r = $\sqrt{2}$, T_n = $(\sqrt{2})^{n + 1}$, T₁₁ = 64.