Show that the progression -27, 9, -3, 1, $-\dfrac{1}{3}$, …… is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 9th term.

Given,

-27, 9, -3, 1, $-\dfrac{1}{3}$,……

$\Rightarrow \dfrac{9}{-27} = \dfrac{-3}{9} = -\dfrac{1}{3}$.

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

a = -27

r = $\dfrac{9}{-27} = -\dfrac{1}{3}$

We know that,

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

T_n = ar^{n - 1}

$\Rightarrow T_n = -27 \times \Big(-\dfrac{1}{3}\Big)^{n - 1} \\[1em] = (-3)^3 \times \Big(-\dfrac{1}{3}\Big)^{n - 1} \\[1em] = (-3)^3 \times \dfrac{1}{(-3)^{n - 1}} \\[1em] = (-3)^{3 -(n - 1)} \\[1em] = (-3)^{3 -n + 1} \\[1em] = (-3)^{4 - n} \\[1em] = \dfrac{1}{(-3)^{n - 4}}.$

9th term

$\Rightarrow T_9 = \dfrac{1}{-3^{9 - 4}} \\[1em] = \dfrac{1}{-3^{5}} \\[1em] = -\dfrac{1}{243} \\[1em]$

Hence, a = -27, r = $-\dfrac{1}{3}$, T_n = = $\dfrac{1}{(-3)^{n - 4}}$, T₉ = $-\dfrac{1}{243}$.