Show that the progression 0.4, 0.8, 1.6,….. is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 7th term.

Given,

0.4, 0.8, 1.6,…..

$\Rightarrow \dfrac{0.8}{0.4}= {\dfrac{1.6}{0.8}} = 2.$

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

a = 0.4

r = $\dfrac{0.8}{0.4}$ = 2

We know that,

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

T_n = ar^{n - 1}

$T_n = 0.4(2)^{n - 1} \\[1em] = \dfrac{4}{10}(2)^{n - 1} \\[1em] = \dfrac{2 \times 2}{10}(2)^{n - 1} \\[1em] = \dfrac{2}{5}(2)^{n - 1} \\[1em] = \dfrac{2^{n + 1 - 1}}{5} \\[1em] = \dfrac{2^{n}}{5}.$

7th term,

T₇ = $\dfrac{2^{7}}{5}$

= $\dfrac{128}{5}$.

Hence, a = 0.4, r = 2, T_n = $\dfrac{2^{n}}{5}$, T₇ = $\dfrac{128}{5}$.