In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.

Let nth term be the last term of the G.P.

Given, a = 7, l = a_n = 448 and S_n = 889.

Using formula,

$a_n = ar^{n - 1} \\[1em] \Rightarrow 448 = 7(r)^{n - 1} \\[1em] \Rightarrow r^{n - 1} = 64 \\[1em] \Rightarrow \dfrac{r^n}{r} = 64 \\[1em] \Rightarrow r^n = 64r. \text{ (Eq 1)} \\[1em]$

Given, S_n = 889

Putting value of rⁿ from Eq 1 in the formula $S_n = \dfrac{a(r^n - 1)}{r - 1}$ we get,

$\Rightarrow 889 = \dfrac{7(64r - 1)}{r - 1} \\[1em] \Rightarrow 889(r - 1) = 448r - 7 \\[1em] \Rightarrow 889r - 889 = 448r - 7 \\[1em] \Rightarrow 889r - 448r = 889 - 7 \\[1em] \Rightarrow 441r = 882 \\[1em] \therefore r = 2.$

Hence, the common ratio of the G.P. is 2.