Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728.

Given, r = 3, l = 486 and S_n = 728.

Let the last term be nth term.

Hence, l = a_n = 486.

Using formula,

$a_n = ar^{n - 1} \\[1em] \Rightarrow 486 = a(3)^{n - 1} \\[1em] \Rightarrow a\dfrac{3^n}{3} = 486 \\[1em] \Rightarrow 3^na = 486 \times 3 \\[1em] \Rightarrow 3^na = 1458 \\[1em] \Rightarrow a = \dfrac{1458}{3^n}. \text{ (Eq 1)}$

Given, $S_n = 728$,

Putting value of $a = \dfrac{1458}{3^n}$ in formula $S_n = \dfrac{a(r^n - 1)}{r - 1}$ we get,

$\Rightarrow S_n = \dfrac{\dfrac{1458}{3^n}(3^n - 1)}{3 - 1} \\[1em] \Rightarrow 728 = \dfrac{1458 - \dfrac{1458}{3^n}}{2} \\[1em] \Rightarrow 1458 - \dfrac{1458}{3^n} = 728 \times 2 \\[1em] \Rightarrow \dfrac{1458}{3^n} = 1458 - 1456 \\[1em] \Rightarrow \dfrac{1458}{3^n} = 2 \\[1em] \Rightarrow 3^n = \dfrac{1458}{2} \\[1em] \Rightarrow 3^n = 729$

Putting the above value in Eq 1,

$\Rightarrow a = \dfrac{1458}{729} \\[1em] \therefore a = 2.$

Hence, the first term of the G.P. is 2.