Question

The sum of a G.P. whose common ratio is $3$ is $728$, and the last term is $486$; find the first term.

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

$2$

$a,ar,a{r}^2$ , $a{r}^3$ , ........................$a{r}^{n-1}$ be the G. P.
$S_n$ be sum first $n$ terms of G.P
Let common ratio $=r$
and let $a_n$ be the nth term of G.P
Given that $r=3$, last term i.e $a_n=a{r}^{n-1}=486$
$S_n=\frac{a(r^n-1)}{r-1}=\frac{ra{r}^{n-1}-a}{r-1}=728$
Substituting the vlues in above eqn.
$\frac{\left(3\right)\times 486-a}{3-1}=728$
$\left(3\right)\times 486-2\times 728=a$
First term i.e. $a=2$