If the fourth and ninth terms of a G.P. are 54 and 13122 respectively, find the G.P. Also, find its general term.

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a₄ = 54

⇒ ar³ = 54 ……..(i)

Also,

⇒ a₉ = 13122

⇒ ar⁸ = 13122 ………(ii)

Dividing (ii) by (i) we get,

$\Rightarrow \dfrac{ar^8}{ar^3} = \dfrac{13122}{54} \\[1em] \Rightarrow r^5 = 243 \\[1em] \Rightarrow r^5 = 3^5 \\[1em] \Rightarrow r = 3.$

Substituting r in (i) we get,

$\Rightarrow a(3)^3 = 54 \\[1em] \Rightarrow 27a = 54 \\[1em] \Rightarrow a = 2.$

n^th term of a G.P. = ar^{n - 1}

= 2(3)^{n - 1}

= 2 × 3^{n - 1}

2^nd term of a G.P. = 2 × 3^{2 - 1}

= 2 × 3

= 6.

3^rd term of a G.P. = 2 × 3^{3 - 1}

= 2 × 3²

= 18.

G.P. = 2, 6, 18, 54, ………

Hence, G.P. = 2, 6, 18, 54, ……… and general term = 2 × 3^{n - 1}