Question

The sum of a GP with common ratio $3$ is $364$, and last term is $243$, then the number of terms is

• A. $6$
• B. $5$
• C. $4$
• D. $10$

Answer 1

The correct option is A

$6$

Explanation for the correct option

Given: sum of GP is $364$ having common ratio $r$ is $3$ and last term $a_n$ is $243$

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

$$\begin{gathered} \Rightarrow 243=a{\left(3\right)}^{n-1}\Rightarrow 243=a\frac{\left(3^n\right)}{3} \\ \Rightarrow 243\times 3=3^{n}a \\ \Rightarrow 729=3^{n}a \end{gathered}$$

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

Rewriting and substituting the values,

$$\begin{gathered} 364=a\left(\frac{r^n-1}{r-1}\right) \\ \Rightarrow 364=\frac{a{3}^n-a}{3-1} \\ \Rightarrow 364=\frac{729-a}{2} \\ \Rightarrow 728=729-a \\ \Rightarrow a=1 \end{gathered}$$

Substituting the value of $a$ in the formula for the general term,

$\begin{array}{cc}\Rightarrow & 243=1{\left(3\right)}^{n-1}\\ \Rightarrow & 243=\frac{3^n}{3}\\ \Rightarrow & 3^n=729\\ \Rightarrow & 3^n=3^6\end{array}$

by comparing the base and exponent of both sides

$n=6$

Hence, option (A) is correct.