Find the 8^th term of the sequence :

$\dfrac{3}{4}, 1\dfrac{1}{2}, 3, ………$

Sequence = $\dfrac{3}{4}, \dfrac{3}{2}, 3, ……..$

Calculating ratio between terms,

$\dfrac{\dfrac{3}{2}}{\dfrac{3}{4}} = \dfrac{3 \times 4}{2 \times 3} = 2, \dfrac{3}{\dfrac{3}{2}} = 2$.

Since,

$\dfrac{\dfrac{3}{2}}{\dfrac{3}{4}} = \dfrac{3}{\dfrac{3}{2}} = 2$.

Hence, the sequence is a G.P. with r = 2 and a = $\dfrac{3}{4}$

We know that n^th term of G.P.,

a_n = ar^{n - 1}

a₈ = $\dfrac{3}{4} \times (2)^{8 - 1}$

= $\dfrac{3}{4} \times 2^7$

= $\dfrac{3}{4} \times 128$

= 3 × 32

= 96.

Hence, a₈ = 96.