In a G.P., the third term is 24 and 6th term is 192. Find the 10th term.

Given, a₃ = 24, a₆ = 192.

By formula, a_n = ar^{n - 1}.

⇒ a₃ = ar^{(3 - 1)}
⇒ ar² = 24.             (Eq 1)

⇒ a₆ = ar^{(6 - 1)}
⇒ ar⁵ = 192.            (Eq 2)

Dividing Eq 2 by Eq 1

$\Rightarrow \dfrac{ar^5}{ar^2} = \dfrac{192}{24} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = (2)^3 \\[1em] \Rightarrow r = 2$

Putting value of r in Eq 1,

$\Rightarrow a(2)^2 = 24 \\[1em] \Rightarrow 4a = 24 \\[1em] \Rightarrow a = 6 \\[1em] \text{10th term of G.P.} = a${10} \\[1em] \Rightarrow a{10} = 6(2)^{10 - 1} \\[1em] = 6(2)^9 = 6 \times 512 = 3072.

Hence, the 10th term of the G.P. is 3072.