Show that the progression $-\dfrac{3}{4}, \dfrac{1}{2}, -\dfrac{1}{3}, \dfrac{2}{9}$,….. is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 6th term

Given,

$-\dfrac{3}{4}, \dfrac{1}{2}, -\dfrac{1}{3}, \dfrac{2}{9}$,…..

$\Rightarrow \dfrac{\dfrac{1}{2}}{-\dfrac{3}{4}} = \dfrac{-\dfrac{1}{3}}{\dfrac{1}{2}} = -\dfrac{2}{3}.$

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

a = $-\dfrac{3}{4}$

r = $\dfrac{\dfrac{1}{2}}{-\dfrac{3}{4}} = \dfrac{-4}{6} = \dfrac{-2}{3}$.

We know that,

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

T_n = ar^{n - 1}

$\Rightarrow T_n = -\dfrac{3}{4} \cdot \Big(-\dfrac{2}{3}\Big)^{n - 1} \\[1em] = \dfrac{-3}{2^2} \cdot \Big(\dfrac{2^{n - 1}}{(-3)^{n - 1}}\Big) \\[1em] = \Big(\dfrac{2^{n - 1 - 2}}{(-3)^{n - 1 - 1}}\Big) \\[1em] = \Big(\dfrac{2^{n - 3}}{(-3)^{n - 2}}\Big)$

6th term,

$T_6 = \Big(\dfrac{2^{6 - 3}}{(-3)^{6 - 2}}\Big) \\[1em] = \Big(\dfrac{2^3}{(-3)^4}\Big) \\[1em] = \dfrac{8}{81}.$

Hence, a = $-\dfrac{3}{4}$, r = $-\dfrac{2}{3}$, T_n = $\Big(\dfrac{2^{n - 3}}{(-3)^{n - 2}}\Big)$, T₆ = $\dfrac{8}{81}$.