**Question:**

The common ratio of a finite GP is 3, and its last term is 486. If the sum of these terms is 728, find the first term.

**Solution:**

'Tn' represents the $\mathrm{n}^{\text {th }}$ term of a G.P. series.

$T_{n}=a r^{n-1}$

$\Rightarrow 486=a(3)^{n-1}$

$\left.\Rightarrow 486=a\left(3^{n} \div 3\right)\right)$

$\Rightarrow 486 \times 3=a\left(3^{n}\right)$

$\Rightarrow 1458=a\left(3^{n}\right) \ldots \ldots \ldots .(i)$

Sum of a G.P. series is represented by the formula, $\mathrm{Sn}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$ when r≠1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

$\therefore 728=a \times \frac{3^{n}-1}{3-1}$

$\Rightarrow 728=a \times \frac{3^{n}-1}{2}$

$\Rightarrow 728 \times 2=a\left(3^{n}\right)-a \ldots \ldots$ [Putting $a\left(3^{n}\right)=1458$ from (i)]

$\Rightarrow 1456=1458-a$

$\Rightarrow 1456-1458=-a$

$\Rightarrow-2=-a$ [Multipying both sides by $-1$ ]

⇒a = 2