If the third, sixth and the last terms of a G.P. are 6, 48 and 3072 respectively, find the first term and the numbers of terms in the G.P.

Let first term be a, common ratio be r and n be number of terms.

Given,

Third, sixth and the last terms of a G.P. are 6, 48 and 3072 respectively.

⇒ ar² = 6 ………(1)

⇒ ar⁵ = 48 ……….(2)

⇒ ar^{n - 1} = 3072 ………..(3)

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{ar^5}{ar^2} = \dfrac{48}{6} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = 2^3 \\[1em] \Rightarrow r = 2.$

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

⇒ a(2)² = 6

⇒ 4a = 6

⇒ a = $\dfrac{6}{4} = \dfrac{3}{2}$.

Substituting value of a in equation (3), we get :

$\Rightarrow \dfrac{3}{2} \times (2)^{n - 1} = 3072 \\[1em] \Rightarrow 3 \times 2^{n - 2} = 3072 \\[1em] \Rightarrow 2^{n - 2} = \dfrac{3072}{3} \\[1em] \Rightarrow 2^{n - 2} = 1024 \\[1em] \Rightarrow 2^{n - 2} = 2^{10} \\[1em] \Rightarrow n - 2 = 10 \\[1em] \Rightarrow n = 12.$

Hence, first term = $\dfrac{3}{2}$ and number of terms in the G.P. = 12.