The third term of a G.P. is greater than its first term by 9 whereas its second term is greater than the fourth term by 18. Find the G.P.

Let first term be a and common ratio be r.

By formula,

⇒ a_n = ar^{n - 1}

Given,

The third term of a G.P. is greater than its first term by 9.

∴ a₃ - a = 9

⇒ ar^{3 - 1} - a = 9

⇒ ar² - a = 9

⇒ a(r² - 1) = 9

⇒ a = $\dfrac{9}{r^2 - 1}$ …….(1)

Given,

Second term is greater than the fourth term by 18.

∴ a₂ - a₄ = 18

⇒ ar^{2 - 1} - ar^{4 - 1} = 18

⇒ ar - ar³ = 18

⇒ ar(1 - r²) = 18

Substituting value of a from equation (1) in above equation, we get :

$\Rightarrow \dfrac{9}{r^2 - 1} \times r(1 - r^2) = 18 \\[1em] \Rightarrow \dfrac{9}{r^2 - 1} \times -r(r^2 - 1) = 18 \\[1em] \Rightarrow -9r = 18 \\[1em] \Rightarrow r = -\dfrac{18}{9} = -2.$

Substituting value of r in equation (1), we get :

$\Rightarrow a = \dfrac{9}{(-2)^2 - 1} \\[1em] = \dfrac{9}{4 - 1} \\[1em] = \dfrac{9}{3} \\[1em] = 3.$

G.P. = a, ar, ar², ar³, …….

= 3, 3 × -2, 3 × (-2)², 3 × (-2)³, …….

= 3, -6, 3 × 4, 3 × -8, ……..

= 3, -6, 12, -24, ……..

Hence, required G.P. = 3, -6, 12, -24, ……..