The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.

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

Given,

⇒ a₄ = 10

⇒ ar³ = 10 ……..(i)

Also,

⇒ a₇ = 80

⇒ ar⁶ = 80 ………(ii)

Dividing (ii) by (i) we get,

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{80}{10} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = \sqrt[3]{8} \\[1em] \Rightarrow r = 2.$

Substituting r in (i) we get,

$\Rightarrow a(2)^3 = 10 \\[1em] \Rightarrow 8a = 10 \\[1em] \Rightarrow a = \dfrac{10}{8} \\[1em] \Rightarrow a = \dfrac{5}{4}.$

Let n be no. of terms,

ar^{n - 1} = 2560

$\Rightarrow \dfrac{5}{4}\times (2)^{n - 1} = 2560 \\[1em] \Rightarrow (2)^{n - 1} = \dfrac{4}{5} \times 2560 \\[1em] \Rightarrow (2)^{n - 1} = 4 \times 512 \\[1em] \Rightarrow (2)^{n - 1} = 2048 \\[1em] \Rightarrow (2)^{n - 1} = (2)^{11} \\[1em] \Rightarrow n - 1 = 11 \\[1em] \Rightarrow n = 12.$

Hence, first term = $\dfrac{5}{4}$, common ratio = 2 and number of terms = 12.