A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.

Let n^th term be the last term.

⇒ ar^{n - 1} = 486

⇒ a(3)^{n - 1} = 486

⇒ a.3ⁿ.3^-1 = 486

⇒ $3^n = \dfrac{486 \times 3}{a} = \dfrac{1458}{a}$

Since, r > 1

$\Rightarrow S = \dfrac{a(r^n - 1)}{(r - 1)} \\[1em] \Rightarrow 728 = \dfrac{a \times (3^n - 1)}{3 - 1} \\[1em] \Rightarrow 728 = \dfrac{a \times \Big(\dfrac{1458}{a} - 1\Big)}{2} \\[1em] \Rightarrow 728 = \dfrac{a \times (1458 - a)}{2a} \\[1em] \Rightarrow \dfrac{1458 - a}{2} = 728 \\[1em] \Rightarrow 1458 - a = 1456 \\[1em] \Rightarrow a = 1458 - 1456 \\[1em] \Rightarrow a = 2.$

Hence, first term = 2.