Question

Find the 15^th term of the G.P. 3, 12, 48, 192, ...

Answer 1

In the given G.P., we have t₁ = 3, t₂ = 12, t₃ = 48 and t₄ = 192.
Thus, we have the first term, i.e., a = t₁ = 3, and the common ratio, i.e., $r=\frac{t_2}{t_1}=\frac{12}{3}=4$.
We know that the n^th term of the G.P. is t_n = ar^{n – 1}.
By taking a = 3, r = 4 and n = 15, we have:
t₁₅ = (3)(4)^{15 – 1}
$\Rightarrow$t₁₅ = (3)(4)¹⁴
Thus, the 15^th term is (3)(4)¹⁴.