Question

The sum of a few terms of any ratio series is $728$, if the common ratio is $3$ and the last term is $486$, then the first term of the series will be

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

Compute the first term.

Given: sum of $n$terms of a GP is $728$, common ratio $\left(r\right)$ is $3$ and the last term$\left(a_n\right)$ is $486$.

Last term, $a_n=a{r}^{n-1}$

Sum of a GP, $S_n=a\frac{r^n-1}{r-1}$

$\Rightarrow$ $\begin{array}{rcl}728& =& \frac{a·r^{n-1}·r-a}{r-1}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}728& =& \frac{486·3-a}{3-1}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}728\times 2& =& 1458-a\end{array}$

$\Rightarrow$ $\begin{array}{rcl}1456& =& 1458-a\end{array}$

$\Rightarrow$ $\begin{array}{rcl}a& =& 1458-1456\end{array}$

$\Rightarrow$ $a=2$

Thus, the first term of the series will is $2$.

Hence, option $A$ is the correct answer.